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probability of rolling 3 dice numbers which product is a multiple of 6
As the question states, I need to find the probability of rolling 3 dice numbers which product is a multiple of 6. what I have tried:
$$\omega={(6,k,k), (3,2,k), (3,4,k))}$$
$$P(A) = \frac{3\times(1^1\times6^2) + 6\times(1^1\times1^1\times6^1) + 6\times(1^1\times1^1\times6^1)}{6^3}=\frac{180}{216}=\frac{5}{6}$$
I know writing $$1^1$$ seems silly, but just trying to show what I'm doing
Thanks for helping and sorry if it is a bad question
• This looks wrong at a glance as it looks like you are overcounting. For instance, where does the outcome where all three dice show a $6$ get counted and how many times did you count it? It should only have been counted once, but it looks like you counted it three times. What about the outcome where you rolled a $2,3,6$? in some order? Was it counted in the first category? or in the second? Jun 14 at 14:41
• You have overcounted since there is an overlap between the cases you describe. E.g. (6,2,3) is in two of the cases so is counted twice. Jun 14 at 14:42
• I think it would be easier to count the outcomes where the product is not even, count the outcomes where the product is not a multiple of three, and count the outcomes where the product is neither even nor a multiple of three, and use that to proceed to a solution. Jun 14 at 14:43
• @JMoravitz what if i use the same way i did but substract the cases where i overcounted? Jun 14 at 14:47
• That seems tedious. The question is crafted in such a way as to specifically suggest using inclusion-exclusion and basic facts about divisibility to approach. If you want to explore other additional approaches after the fact, go ahead and revisit the problem, but I strongly encourage you to use the approach I suggested in my second comment first. Jun 14 at 14:48
Let the dice be thrown in sequence or otherwise be distinguishable some other way. There are $$6^3 = 216$$ total different equally likely outcomes. | {
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Approach by trying to count how many of these outcomes were "bad" and should be removed because they either are not multiples of $$2$$, are not multiples of $$3$$, and keeping in mind that they might have been both.
Of these, $$3^3=27$$ are "bad" because their product of the results is not even (and thus not a multiple of six), seen by noting a product is odd iff all terms in the product is odd, there being three odd numbers.
$$~$$
Similarly, $$4^3=64$$ are "bad" because their product of results is not a multiple of three (and thus not a multiple of six), seen by noting a product is not a multiple of three iff all terms in the product are not multiples of three, there being four possible non-multiple of three numbers to choose from here.
$$~$$
Finally, some of these "bad" outcomes we counted twice as they were simultaneously not even and not multiples of three. These would be the outcomes consisting only of $$1$$'s and $$5$$'s, there being $$2^3=8$$ of which. We keep this in mind to correct our count as we proceed with inclusion-exclusion.
The probability then is:
$$\frac{6^3-3^3-4^3+2^3}{6^3} = \frac{216 - 27 - 64 + 8}{216} = \frac{133}{216}$$
Let $$a_2,b_2,c_2$$ ($$a_3,b_3,c_3$$) denote the exponents of $$2$$ ($$3$$) in the prime factorization of the number shown on the three dice. Then the probability you want is
$$P(a_2+b_2+c_2>0, a_3+b_3+c_3>0) = 1-P(a_2+b_2+c_2=0\ or \ a_3+b_3+c_3=0)$$
$$= 1-P(a_2+b_2+c_2=0)-P(a_3+b_3+c_3=0)+P(a_2+b_2+c_2=0, a_3+b_3+c_3=0)$$
$$(by\ independence)= 1-P(a_2=0)^3-P(a_3=0)^3+P(a_2=0,a_3=0)^3$$
$$= 1-P(1,3,\ or \ 5)^3-P(1,2,4, \ or \ 5)^3+P(1 \ or \ 5)^3$$
$$= 1 - (\frac{3}{6})^3 - (\frac{4}{6})^3+(\frac{2}{6})^3 = \frac{133}{216}.$$
• very elegant approach ! Jun 14 at 15:32 | {
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# Which matrices are covariances matrices?
Let $V$ be a matrix.
What conditions should we require so that we can find a random vector $X = (X_1, \dots, X_n)$ so that $V = Var(X)$?
Of course necessary conditions are:
• All the elements on the diagonal should be positive
• The matrix has to be symmetric
• $v_{ij} \le \sqrt{v_{ii}v_{jj}}$ (Because of $Cov(X_i, X_j) \le \sqrt{Var(X_i) Var(X_j)})$
But I am sure these are not sufficient as I have a counterexample.
So what other properties we should require on a matrix so that it can be considered a covariance matrix?
I think I cleared this up sufficiently.
Okay, so
1) If $V$ is not semi definite positive, then such a vector $X$ does not exists. (Since all covariances matrix are semi definite positive)
2) If $V$ is symmetric semidefinite positive, then such an $X$ exists! [0]
This implies that
$$\text{exists a random vector X: V = Cov(X)} \iff \text{V is symmetric positive semidefinite}$$
Since we know that those I listed in the question are necessary condition for $V$, we deduce that all symmetric semidefinite positive matrices have elements on the diagonal $\ge 0$ and are such that $v_{ij} \le \sqrt{v_{ii}v_{jj}}$.
These are not sufficient though for a matrix to be semidefinite positive but sufficient conditions are well known, after all.
[0] Proof
Since $V$ is symetric is possible to find an orthogonal matrix $Q$ such that $V = QDQ^T$, where $D$ is a diagonal matrix whose values are the eigenvalues of $V$. If $V$ is semipositive definite the elements of $D$ are all $\ge 0$, hence we can find $X$ such that $D = Cov(X)$ (just take all the variables independent with the specified variance)
It follows that the random vector $QX$ has covariance equal to $$Cov(QX) = QCov(X)Q^T = QDQ^T = V$$ | {
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It follows that the random vector $QX$ has covariance equal to $$Cov(QX) = QCov(X)Q^T = QDQ^T = V$$
• There's something unclear about your proof: where exactly did you use the condition $v_{ij} \leq \sqrt{v_{ii}v_{jj}}$? Is this a consequence of the positive semidefinite/symmetric matrix condition? Beacuse if not, you there's something wrong in your conclusion. – RandomGuy Oct 31 '16 at 14:32
• @RandomGuy I didn't. I showed that if a matrix is symmetric, positive semidefinite, then it can be written as $Cov(X)$; therefore it must hold that $v_{ij} \le \sqrt{v_{ii}v_{j}}$ for every symmetric semidefinite positive matrix, because it holds for $Cov(X)$. You can consider this a proof of this statement. Of course, if the statement is false, then my proof is invalid :) – Ant Nov 1 '16 at 20:04
• yes, now it makes sense. My question was about the fact that the statement of your original question included that inequality as a condition and in the answer you answer your own question, which seemed to imply that you were using that condition as a hypothesis. – RandomGuy Nov 2 '16 at 8:18 | {
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# Area of Overlapping Quadrilateral Tiles
I was studying for some quizzes when I stumbled across this question. It goes like this:
Two regular quadrilaterals vinyl tiles each of $1$ foot long on each sides overlap each other such that the overlapping region is a regular octagon. What is the area of the overlapping region?
My work:
I imagined the problem like this:
The area of the shaded blue region is what I want to get. I don't know if one side of the octagon is one-third of that of the side of square, because my calculated area is slightly larger than that of my book's answer.
How do you get the area of the blue region above?
Let $a$ be a length-side of the octagon $ABCDE...$, where $BC$ placed on the square side.
Thus, since $\sin45^{\circ}=\frac{1}{\sqrt2}$, we obtain:$$\frac{1}{2}-\frac{a}{2}=a\cdot\frac{1}{\sqrt2},$$ which gives $a=\sqrt2-1$.
Thus, $\frac{1}{2}-\frac{a}{2}=\frac{1}{2}-\frac{\sqrt2-1}{2}=\frac{2-\sqrt2}{2}$ and the needed area is $$1-4\cdot\frac{1}{2}\left(\frac{2-\sqrt2}{2}\right)^2=1-\frac{1}{2}(2-\sqrt2)^2=2\sqrt2-2$$
• How did you derived the expression above? What is the thought process? – Palautot Ka Aug 23 '17 at 15:16
• @Palautot Ka Say me please, are my answer and the answer in your book, same? – Michael Rozenberg Aug 23 '17 at 15:35
• The area of a regular polygon is $A = \frac{a^2 n }{4} \cot \frac{180^o}{a}.$ Using your $a = \sqrt{2} -1$ and $n = 8$ sides, I got the area to be 0.8284 square feet. Is your value $4 \sqrt{2} - 5$ the area of the blue octagon? – Palautot Ka Aug 23 '17 at 15:35
• @Palautot Ka I fixed my post. See now please. – Michael Rozenberg Aug 23 '17 at 15:39
• The book's answer is 119.29 square inchesXD – Palautot Ka Aug 23 '17 at 15:40 | {
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The area of the shaded region will be an octagon - one square minus 4 white triangles. Say the legs of the right triangles are l. Then $$2l+{\sqrt{2l^2}} = 1$$ from the Pythagorean theorem. Solving for l we get $$1-{\sqrt2}/2$$ The area will then be $$1-4*l^2/2$$ giving us $$2*{\sqrt2}-2$$
• I like the insight...... – Palautot Ka Aug 23 '17 at 15:45
The height of a triangle is the half of the diagonal of a square minus the side of a square.
The area of a triangle (isoceles rectangle) is the square of the height.
Hence the blue area
$$1^2-4\frac{\left(\sqrt2-1\right)^2}4=2\sqrt2-2.$$ | {
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How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational? [duplicate]
So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:
Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.
I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.
I also figured maybe I should play with it this way:
$2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$
But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.
So what am I missing (other than sleep and food)? What route should I take to prove this? Thanks in advance!
marked as duplicate by Watson, Chinnapparaj R, user10354138, Brahadeesh, KReiserNov 26 '18 at 8:38
• – punctured dusk Apr 7 '15 at 12:22
• – Watson Nov 25 '18 at 19:45
• – Watson Nov 25 '18 at 21:45
• – Watson Jan 29 at 9:06
Note
$$1 + \sqrt[3]{2} + \sqrt[3]{4} = \frac{(\sqrt[3]{2})^3 - 1}{\sqrt[3]{2} - 1} = \frac{1}{\sqrt[3]{2} - 1}.$$
So if $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, then $1/(\sqrt[3]{2} - 1)$ is rational, which implies $\sqrt[3]{2} - 1$ is rational. Then $\sqrt[3]{2}$ is rational, a contradiction. | {
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• Very nice answer. +1 – Timbuc Mar 26 '15 at 18:00
• Thanks for the reply. I must be tired or confused because I'm missing something very basic and don't realize how $1 + \sqrt[3]{2} + \sqrt[3]{4} = \frac{(\sqrt[3]{2})^3 - 1}{\sqrt[3]{2} - 1}$. I apologize for my stupidity - it's been a tough day. – Elad Avron Mar 26 '15 at 18:01
• Let $x = \sqrt[3]{2}$. Use the factorization $(1 + x + x^2)(x - 1) = x^3 - 1$ to obtain the identity. – kobe Mar 26 '15 at 18:04
• Of course, silly me. Thank you so much, this solution is brilliant! – Elad Avron Mar 26 '15 at 18:09
• @Elad One easly proves a much more general result: for irrational cube roots of rationals, said property fails only for $\,\sqrt[3]1,\,$ see my answer. $\ \$ – Bill Dubuque Mar 26 '15 at 20:21
$y = \sqrt[3]{2} + \sqrt[3]{4}$ is a root of the equation $y^3 - 6y - 6 = 0$.
(To see this, let $x = \sqrt[3]{2}$ and $y = \sqrt[3]{2}+\sqrt[3]{4}=x+x^2$. Then $y^3 = x^3 + 3x^4 + 3x^5 + x^6 = 2 + 6x + 6x^2 + 4 = 6 + 6y$.)
By the rational root theorem, we know that any rational roots of $y^3 - 6y - 6 = 0$ would have to be in the set $\{\pm1,\pm2,\pm3,\pm6\}$; we can quickly confirm that none of these is in fact a root, meaning that the equation does not have any rational roots.
Therefore, $y$ must be irrational.
• Short, sweet, and intuitive. Very nice. – Jon Mar 26 '15 at 21:16
An easy approach: If $p=\sqrt[3]{2}+\sqrt[3]{4}$ is rational:
$$p^2 = \sqrt[3]{4}+2\cdot 2+ 2\sqrt[3]{2} = p+4+\sqrt[3]{2}.$$
So $\sqrt[3]{2}=p^2-p-4$ would be rational.
An alternative, more general approach.
Claim: If $a,b$ are integers that are not perfect cubes, and $a\neq -b$, then $\sqrt[3]{a}+\sqrt[3]b$ is irrational.
Proof:
Assume $\sqrt[3]{a}+\sqrt[3]{b}$ is rational. Cube it and get:
$$(\sqrt[3]{a}+\sqrt[3]{b})^3 = a+ 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) + b$$
Now, since $a,b$ are rational and $\sqrt[3]{a}+\sqrt[3]{b}$ is non-zero and rational, this means that $\sqrt[3]{ab}$ is rational. | {
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Letting $p=\sqrt[3]{a}+\sqrt[3]{b}$ and $q=\sqrt[3]{ab}$, this means that
$$(x-\sqrt[3]{a})(x-\sqrt[3]{b}) = x^2-px+q$$ is a rational polynomial. It shares at least one root with $x^3-b$, But the GCD of these two polynomials has to be a rational polynomial, so the GCD cannot be linear (since it would be $x-\sqrt[3]b$, which is not a rational polynomial.)
This means that $x^2-px+q$ has to divide $x^3-b$. That means that $\sqrt[3]{a}$ is a root of $x^3-b$, which means that $a=b$. But there are no repeated roots of $x^3-b$, which yields a contradiction.
Corollary: If $a,b$ are rationals such that $a\neq -b$ and $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are irrational, then $\sqrt[3]{a}+\sqrt[3]{b}$ is irrational.
Proof: Rationalize the denominators and revert to the above theorem for integers.
It holds true for any irrational $\,x=\sqrt[3]n,\,$ except $\,n=1\,$ (so $\,x^2+x=-1\in\Bbb Q)$
More generally: if $\ r\in\Bbb Q\$ and $\,x=\sqrt[3]r\not\in\Bbb Q\,$ then $\,x^2+x = q\in\Bbb Q\iff r = 1.$
Proof $\,\ qx = x^3+x^2 = r+x^2\$ so $\ qx-r = x^2 = q-x,\$ so $\,(\color{#c00}{q\!+\!1})\,x = r+q.$
Therefore $\,\ x\not\in\Bbb Q\,\Rightarrow\,\color{#c00}{q = -1}\,\Rightarrow\, 0 = (\color{#c00}{q\!+\!1})x = r+q = r-1,\$ thus $\,\ r = 1.$
• Um, if $n=1$, isn't it $x^2+x=2$? And is $x=\sqrt[3]{1}$ irrational? I'm confused. – Thomas Andrews Mar 29 '15 at 4:54
• @Thomas The hypothesis is that $\,x\,$ is an irrational cube root of $\,n.\ \$ – Bill Dubuque Mar 29 '15 at 6:41
• Well, $\sqrt[3]{n}$ is a single-valued function, by convention. You could say any irrational $x$ with $x^3=n$, I suppose. – Thomas Andrews Mar 29 '15 at 15:25
• @Thomas Algebraists not too infrequently overload the notation to denote any root. I will change it to avoid possible confusion. Thanks for pointing that out. – Bill Dubuque Mar 29 '15 at 16:25 | {
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Here is another approach which generalises to many similar examples, and which involves no complicated simplifications of surds. (In fact, no easy simplifications of surds either.)
First, $\sqrt[3]2$ is an algebraic integer, that is, it is a root of $$x^3-2$$ which is a monic polynomial (leading coefficient $1$) with integer coefficients. Similarly, $\sqrt[3]4$ is an algebraic integer.
It is known that the sum of two algebraic integers is an algebraic integer. Thus, $x=\sqrt[3]2+\sqrt[3]4$ is an algebraic integer.
It is also known that if an algebraic integer is rational, then it is a rational integer - that is, an ordinary integer, $0,1,-1,2,-2$ etc. However, it is not hard to find the estimates $$1<\sqrt[3]2<\frac43 ,\quad \frac43<\sqrt[3]4<\frac53\ ;$$ so $\frac73<x<3$, hence $x$ is not an integer and must be irrational.
If $a$ is rational, $a^2-a-4$ is rational. What is $a^2-a-4$ when $a=\sqrt[3]2+\sqrt[3]4$?
(After you work that out, I bet you'll wonder where I got $a^2-a-4$ from. Or, try to figure it out yourself. Now that you have a sort of idea on how to deal with this sort of problem, I leave you with another, similar problem: What polynomial can I use to prove that $\sqrt[3]3+\sqrt[3]9$ is irrational?)
• I was going to suggest proving $\sqrt[3]2+\sqrt[3]3$ irrational, but then I realized that the polynomial involved is an eight-degree one. And I'm not sure how hard the one I gave is; I didn't try it yet. – Akiva Weinberger Mar 26 '15 at 20:49 | {
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# What's my mistake in finding $\int_0^\infty dx e^{-ax^2} \sin(b/x^2)$?
I want to evaluate $$I=\int_0^\infty dx e^{-ax^2} \sin(b/x^2)$$ for $$a,b>0$$. A first simplification is to substitute $$y=x/\sqrt{a}$$ and define $$c=ab>0$$ to obtain $$I=\frac{1}{\sqrt{a}} \int_0^\infty e^{-x^2} \sin(c/x^2)$$ Now my idea was to use the Taylor series for the sine $$I=a^{-1/2} \int_0^\infty dx e^{-x^2} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\left(\frac{c}{x^2} \right)^{2k+1}$$ Now I interchange sum and integral although I have no justification $$I=a^{-1/2} \sum_{k=0}^\infty \frac{(-1)^kc^{2k+1}}{(2k+1)!} \int_0^\infty dx e^{-x^2} \left(\frac{1}{x^2} \right)^{2k+1}$$ Substituting $$t=x^2$$ in the integral we obtain the gamma function $$I=a^{-1/2} \sum_{k=0}^\infty \frac{(-1)^kc^{2k+1}}{(2k+1)!} \frac{1}{2} \Gamma(-2k-1/2)$$ Using $$\Gamma \left({\frac{1}{2}}-n\right)={(-4)^{n}n! \over (2n)!}{\sqrt {\pi }}$$ (which can be shown using the reflection formula and the duplication formula for the Gamma function) with $$n=2k+1$$ I obtain $$I=a^{-1/2} \sqrt{\pi} \sum_{k=0}^\infty \frac{(-1)^k(-4c)^{2k+1}}{(4k+2)!} \frac{1}{2}$$ or $$I=-\frac{1}{2}\sqrt{\frac{\pi}{a}} \sum_{k=0}^\infty \frac{(-1)^k(4c)^{2k+1}}{(4k+2)!}=-\frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c}) \sinh(\sqrt{2c})$$ where I used wolfram alpha for the last series.
The problem: The above result is wrong. It should be (wolfram alpha and Gradshteyn) $$I=\frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c})\exp(-\sqrt{2c})$$
The question: Can someone spot my mistake? Was it interchanging the limits? I would also be interested in your solutions to the integral $$I$$ using other approaches.
• After the interchange, all the integrals diverge at $0$, aren't they? – user58697 Jul 29 at 19:58
• @user58697 Yes, you are right. Probably that is the mistake... Do you have an idea for an alternative approach to the original integral? – thomasfermi Jul 29 at 20:09 | {
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The Glasser's master theorem is a useful tool for the solution. First, use Euler's formula to decompose the sine term into the sum of exponentials. Then it boils down to computing the integral of the form
$$J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - \frac{p}{x^2} \right) \, \mathrm{d}x.$$
Assume for a moment that $$a, p > 0$$. Then by completing the square, we get
$$J(p) = \int_{0}^{\infty} \exp\left( -a \left( x - \frac{\smash{\sqrt{p/a}}}{x} \right)^2 - 2\sqrt{ap} \right) \, \mathrm{d}x.$$
Then by the Glasser's master theorem and the gaussian integral, this evaluates to
$$J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - 2\sqrt{ap} \right) \, \mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-2\sqrt{ap}). \tag{*}$$
Although $$\text{(*)}$$ is originally proved for $$p > 0$$, both sides of $$\text{(*)}$$ define holomorphic functions for $$p$$ in the right-half plane $$\mathbb{H}_{\to} = \{z \in \mathbb{C} : \operatorname{Re}(z) > 0\}$$ and are continuous on the closed right-half plane $$\overline{\mathbb{H}_{\to}}$$. So by the identity theorem and continuity, $$\text{(*)}$$ extends to all of $$p \in \overline{\mathbb{H}_{\to}}$$. In particular, plugging $$p = \pm ib$$ for $$b > 0$$, we get
$$J(\pm ib) = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-2\sqrt{\pm i ab}) = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-\sqrt{2c}(1\pm i)).$$
Therefore
$$I = \frac{J(-ib) - J(ib)}{2i} = \frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c}) \exp(-\sqrt{2c}).$$
• Thanks for your very nice solution. I had never heard of Glasser's theorem before! – thomasfermi Jul 30 at 22:03 | {
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# help finding sum of series
Printable View
• April 30th 2009, 01:11 AM
adhyeta
help finding sum of series
got 2 series.
1.
12 - 22 + 32 - 42 + ................. 992 - 1002
(ok those are just squares)
2.
1.1! + 2.2! + ....... 50.50!
need to find sum...
• April 30th 2009, 03:07 AM
pickslides
Quote:
Originally Posted by adhyeta
got 2 series.
1.
12 - 22 + 32 - 42 + ................. 992 - 1002
(ok those are just squares)
need to find sum...
$1^2-2^2+3^2-4^2+\cdots+99^2-100^2$
$1-4+9-16+25-36+\cdots+99^2-100^2$
finding the difference in pairs now giving 50 terms.
$-3-7-11-\cdots$
Use sum of an arithmetic sequence
$S_n = \frac{n}{2}(2a+(n-1)d)$
where n = number of terms = 50
a = the first term = -3 and d = the common difference between terms = -4
$S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$
• April 30th 2009, 04:22 AM
Soroban
Hello, adhyeta!
Are you familiar with these summation formulas?
. . $\sum^n_{k=1} 1 \;\;= \;n$
. . $\sum^n_{k=1}k \;\;=\;\frac{n(n+1)}{2}$
Quote:
$1)\;\;S \;=\;1^2 - 2^2 + 3^2 - 4^2 + \hdots + 99^2 - 100^2$
$\text{We have: }\;S \;=\;\underbrace{(1^2 + 3^2 + 5^2 + \hdots + 99^2)}_{\text{first 50 odd squares}} \;-\; \underbrace{(2^2+4^2+6^2 + \hdots + 100^2)}_{\text{first 50 even squares}}$
. . $\text{Then: }S \;=\qquad\qquad \sum^{50}_{k=1}(2k-1)^2 \qquad - \qquad\qquad \sum^{50}_{k=1}(2k)^2$
Hence: . $S \;=\;\sum^{50}_{k=1}\bigg[(2k-1)^2 - (2k)^2\bigg] \;=\;\sum^{50}_{k=1}(1-4k)$
. . $= \;\;\sum^{50}_{k=1}\!1 \;-\; 4\!\sum^{50}_{k=1}k \;\;=\;\;50 - 4\,\frac{50\cdot51}{2} \;\;=\;\;\boxed{-5,\!050}$
• April 30th 2009, 06:10 AM
adhyeta
Quote:
Originally Posted by pickslides
$1^2-2^2+3^2-4^2+\cdots+99^2-100^2$
$1-4+9-16+25-36+\cdots+99^2-100^2$
finding the difference in pairs now giving 50 terms.
$-3-7-11-\cdots$
Use sum of an arithmetic sequence
$S_n = \frac{n}{2}(2a+(n-1)d)$
where n = number of terms = 50
a = the first term = -3 and d = the common difference between terms = -4 | {
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a = the first term = -3 and d = the common difference between terms = -4
$S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$
thats not coming out to be correct...its -5050.
refer to soroban's solution.
• April 30th 2009, 02:11 PM
pickslides
Quote:
Originally Posted by adhyeta
thats not coming out to be correct...its -5050.
refer to soroban's solution.
$
S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))$
$
S_{50} = 25(-6+(49)(-4))$
$
S_{50} = 25(-6+-196)$
$
S_{50} = 25(-202)$
$
S_{50} = -5050$
Worked for me!
• April 30th 2009, 03:04 PM
Jester
Quote:
Originally Posted by adhyeta
got 2 series.
1.
12 - 22 + 32 - 42 + ................. 992 - 1002
(ok those are just squares)
2.
1.1! + 2.2! + ....... 50.50!
need to find sum...
Here's the second one. Let
$S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!$
$= (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n!
$
$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)$
$
= (n+1)! - 1
$
so here $n = 50$ so the answer $S = 51!-1$.
• April 30th 2009, 05:16 PM
adhyeta
Quote:
Originally Posted by pickslides
$
S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))$
$
S_{50} = 25(-6+(49)(-4))$
$
S_{50} = 25(-6+-196)$
$
S_{50} = 25(-202)$
$
S_{50} = -5050$
Worked for me!
hey! so sorry. i think i went wrong with the calc.(-calculation-)
(Happy)
• May 3rd 2009, 07:16 AM
adhyeta
Quote:
Originally Posted by danny arrigo
Here's the second one. Let
$S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!$
$= (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n!
$
$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)$
$
= (n+1)! - 1
$
so here $n = 50$ so the answer $S = 51!-1$.
hey! can we similarily prove the sum $n^{2}.n!$??? | {
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Calculating amplitude, time-shift and phase of third sinusoid as a sum of two others
My apologies in advance, I am new to this digital signal processing. Question is at the bottom
I have two 4000hz sinusoids given by the following formulas: unknown frequency such that: $$x_1(t) = 24\cos(2\pi4000(t-t_{m1})) \\ x_2(t) = 28.8 \cos(2\pi4000(t-t_{m2}))$$ and a third given by $$x_3(t) = x_1(t) + x_2(t)$$ where $$t_{m1} = 0.000775 \\ t_{m2} = -0.00258$$
I have calculated the phase of the first two sinusoids (in radians) as $$\phi_1 = 0.329 \\ \phi_2 = -1.132$$ with the formula $$\phi_i = \omega_if_i$$ where $$\omega_i = 2\pi f$$
Finally, I need to calculate the amplitude $$A_3$$ and phase $$\phi_3$$ of $$x_3(t)$$. I am unsure which formula to use to do so. I believe $$A_3$$ is calculated as: $$a_3 = \sqrt{A_1^2 + A_2^2} = 37.49$$
How would I calculate the time shift and phase angle of $$x_3(t)$$?
• Are you familiar with complex numbers ? – Hilmar Sep 27 '19 at 18:38
• @hilmar I am, but not with respect to this topic. – kp-a Sep 27 '19 at 20:34
Here's how I would do it.
Use the Cosine angle addition formula on $$x_1$$ and $$x_2$$:
$$\cos( \alpha + \beta) = \cos( \alpha ) \cos(\beta)-\sin( \alpha ) \sin(\beta)$$
Like this:
$$x_1(t) = a_1 \cos( \omega t + \phi_1 )$$
$$x_1 = a_1 \cos( \omega t ) \cos(\phi_1)- a_1 \sin( \omega t ) \sin(\phi_1)$$
The $$\omega$$s will be the same, so:
$$x_2 = a_2 \cos( \omega t ) \cos(\phi_2)- a_2 \sin( \omega t ) \sin(\phi_2)$$
You can now rearrange and add them together:
\begin{align} x_3 &= x_1 + x_2 \\ &= \left[a_1 \cos(\phi_1) + a_2 \cos(\phi_2) \right] \cos( \omega t ) - \left[a_1 \sin(\phi_1) + a_2 \sin(\phi_2) \right] \sin( \omega t ) \end{align}
Since you know the $$\phi$$s and $$a$$s, you can put this in the form:
$$x_3 = C \cos( \omega t ) - D \sin( \omega t )$$
Now, you just have to reverse the process to get it back into time phase form.
$$\theta = \operatorname{atan2}(D,C)$$
$$M = \sqrt{ C^2 + D^2 }$$ | {
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$$\theta = \operatorname{atan2}(D,C)$$
$$M = \sqrt{ C^2 + D^2 }$$
Therefore:
$$x_3 = M \cos( \omega t + \theta )$$
$$x_3 = M \cos\left( \omega \left( t + \frac{\theta}{\omega} \right) \right)$$
Now, just plug and chug and you should have it.
The latter part might be better understood in a different order:
\begin{align} x_3(t) &= a_3 \cos( \omega t + \phi_3 )\\ &= a_3 \cos( \omega t ) \cos(\phi_3)- a_3 \sin( \omega t ) \sin(\phi_3) \end{align}
From the equations above:
\begin{align} C &= a_3 \cos(\phi_3)\\ D &= a_3 \sin(\phi_3) \end{align}
From there it follows:
$$a_3 = \sqrt{C^2+D^2}$$
and
$$\phi_3 = \tan^{-1}\left(\frac{D}{C}\right)$$
The thing is, both of these last two equations have two possible solutions each and they need to be matched. Conveniently, the atan2 function is available on most platforms which yields the correct angle from the latter for the positive root of the former.
This solution is actually the proof of a very important principle: When two pure tones of the same frequency are added together the result is a pure tone of the same frequency although it is possible for it to have zero amplitude.
In order for the zero amplitude case to occur, the two tones have to have the same amplitude and be a half cycle ($$\pi$$ radians, 180 degrees) out of phase. This is called complete destructive interference.
If the two tones are in phase, the amplitudes are merely added and the phase remains the same. This is technically called complete constructive interference.
By repeated application of this principle, the same holds true for any linear combination of tones.
This shows up in a surprising number of places.
a. Perhaps the discussion at: https://www.dsprelated.com/showarticle/635.php would be of some value to you. | {
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• You beat me to it, I remember reading this article a few years ago! – Ben Sep 28 '19 at 16:49
• Welcome back to this forum. Looking forward to seeing what you have to say in the next item in your answer (the one that will begin "b.") when you get around to adding it. – Dilip Sarwate Sep 28 '19 at 17:03
• Hi Dilip. (Hope you are doing well.) I don't know what you mean by "the next item in your answer (the one that will begin "b.")." – Richard Lyons Oct 2 '19 at 10:49
• @DilipSarwate, Hey Rick, you want to start your comment as I did if you want the little notification thingy to work. Your post begins: "a. Perhaps..". I, too, have been breathlessly awaiting what might be the "b." – Cedron Dawg Oct 2 '19 at 18:15
• @CedronDog, Hi Cedron. Ah, ...now I see. I don't remember typing the two 'a.' characters in my Answer. So my eyes didn't actually see them in my Answer. Thanks for helping me. In any case Mbaz can obtain his desired results by using the top row of Table 1 in my blog at my above posted web link, or he can use your equations. Happily, your and my equations produce identical results. – Richard Lyons Oct 3 '19 at 10:34 | {
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# Finding cosine when negating its angle, given only the value of sine.
If $$\sin(\theta) = \frac{1}{\sqrt{5}}$$, I must find $$\cos(-\theta)$$.
Using a right triangle (hypotenuse is $$\sqrt{5}$$ and opposite is $$1$$), it is easy to find that the adjacent side is $$2$$. Then since $$\cos(-\theta) = \cos(\theta)$$ I can simply find the answer as $$\frac{2}{\sqrt{5}}$$.
This is correct according to my homework. However, I was thinking, what if $$\theta$$ is in the second quadrant of the unit circle? Basically $$\pi > \theta > \pi/2$$.
Then, visually speaking if you look at the unit circle, there is no way that $$\cos(\theta)$$ can be positive. It has to be negative because the angle ends up in the third quadrant when you negate it as $$-\theta$$.
But my homework says that the only answer is the positive version, so I'm not sure what am I doing wrong.
• If you are only given that $\sin(\theta)=1/\sqrt 5$, there are two possible values for $\cos(\theta)$. It is possible that the answers are incomplete and/or the teacher did not think about negative values. – D.R. Dec 15 '19 at 5:10
I think you have put more good thought into this than the author of your homework answers did, and I agree with your conclusion.
Representing the angle $$\theta$$ on the unit circle, we have one end of the angle at $$(1,0)$$ and the other end at any point where $$y=\frac1{\sqrt5}$$, which is to say the other end can be at either $$\left(\frac2{\sqrt5},\frac1{\sqrt5}\right)$$ or $$\left(-\frac2{\sqrt5},\frac1{\sqrt5}\right)$$.
There are infinitely many angles with sine $$\frac1{\sqrt5}$$, since you can go around the circle as many times as you like in either direction to reach one of those two points, but you must end at one of those two points.
Then the other end of the angle $$-\theta$$ is at $$\left(\frac2{\sqrt5},-\frac1{\sqrt5}\right)$$ or $$\left(-\frac2{\sqrt5},-\frac1{\sqrt5}\right)$$. | {
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In the end there are two possible values for $$\cos(-\theta)$$ found by taking the $$x$$ coordinates of the last two points: $$\frac2{\sqrt5}$$ and $$-\frac2{\sqrt5}.$$
(Or you could realize that $$\cos(-\theta)=\cos(\theta)$$ and therefore when you found the endpoints of the angle $$\theta$$ you already had the answer.)
But if someone did not think about it as much as you did, they might not think about the answer $$-\frac2{\sqrt5}.$$ I guess that's what happened to the author of your homework answers.
A big reason why I was so explicit about using the unit circle in this answer is that it encourages thinking about all the quadrants, whereas the right-triangle definition of sine and cosine really only makes sense in the first quadrant. | {
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# 3.4 Composition of functions (Page 6/9)
Page 6 / 9
## Finding the domain of a composite function involving radicals
Find the domain of
Because we cannot take the square root of a negative number, the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(-\infty ,3\right].\text{\hspace{0.17em}}$ Now we check the domain of the composite function
The domain of this function is $\text{\hspace{0.17em}}\left(-\infty ,5\right].\text{\hspace{0.17em}}$ To find the domain of $\text{\hspace{0.17em}}f\circ g,\text{\hspace{0.17em}}$ we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since $\text{\hspace{0.17em}}\left(-\infty ,3\right]\text{\hspace{0.17em}}$ is a proper subset of the domain of $\text{\hspace{0.17em}}f\circ g.\text{\hspace{0.17em}}$ This means the domain of $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ is the same as the domain of $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ namely, $\text{\hspace{0.17em}}\left(-\infty ,3\right].$
Find the domain of
$\left[-4,0\right)\cup \left(0,\infty \right)$
## Decomposing a composite function into its component functions
In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function , so we may choose the decomposition that appears to be most expedient.
## Decomposing a function
Write $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{5-{x}^{2}}\text{\hspace{0.17em}}$ as the composition of two functions. | {
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We are looking for two functions, $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h,\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}f\left(x\right)=g\left(h\left(x\right)\right).\text{\hspace{0.17em}}$ To do this, we look for a function inside a function in the formula for $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ As one possibility, we might notice that the expression $\text{\hspace{0.17em}}5-{x}^{2}\text{\hspace{0.17em}}$ is the inside of the square root. We could then decompose the function as
We can check our answer by recomposing the functions.
$g\left(h\left(x\right)\right)=g\left(5-{x}^{2}\right)=\sqrt{5-{x}^{2}}$
Write $\text{\hspace{0.17em}}f\left(x\right)=\frac{4}{3-\sqrt{4+{x}^{2}}}\text{\hspace{0.17em}}$ as the composition of two functions.
$\begin{array}{l}g\left(x\right)=\sqrt{4+{x}^{2}}\\ h\left(x\right)=\frac{4}{3-x}\\ f=h\circ g\end{array}$
Access these online resources for additional instruction and practice with composite functions.
## Key equation
Composite function $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
## Key concepts | {
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## Key concepts
• We can perform algebraic operations on functions. See [link] .
• When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
• The function produced by combining two functions is a composite function. See [link] and [link] .
• The order of function composition must be considered when interpreting the meaning of composite functions. See [link] .
• A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
• A composite function can be evaluated from a table. See [link] .
• A composite function can be evaluated from a graph. See [link] .
• A composite function can be evaluated from a formula. See [link] .
• The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See [link] and [link] .
• Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
• Functions can often be decomposed in more than one way. See [link] .
## Verbal
How does one find the domain of the quotient of two functions, $\text{\hspace{0.17em}}\frac{f}{g}?\text{\hspace{0.17em}}$
Find the numbers that make the function in the denominator $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ equal to zero, and check for any other domain restrictions on $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ such as an even-indexed root or zeros in the denominator. | {
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why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
madras university algebra questions papers first year B. SC. maths
Hey
Rightspect
hi
chesky
Give me algebra questions
Rightspect
how to send you
Vandna
What does this mean
cos(x+iy)=cos alpha+isinalpha prove that: sin⁴x=sin²alpha
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
is there any case that you can have a polynomials with a degree of four?
victor
***sscc.edu/home/jdavidso/math/catalog/polynomials/fourth/fourth.html
Oliver
can you solve it step b step
give me some important question in tregnamentry
Anshuman
what is linear equation with one unknown 2x+5=3
-4
Joel
x=-4
Joel
x=-1
Joan
I was wrong. I didn't move all constants to the right of the equation.
Joel
x=-1
Cristian
y=x+1
gary
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function: 1--- g(f(x)) = x for all x in the domain of f 2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x.
Grace | {
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# How is the inductive hypothesis in strong mathematical induction different from that in ordinary induction?
I don't understand how supposing that $$P(k), k\geq 1$$ for ordinary induction is different from $$P(i), 1 \leq i \leq k, k\geq1$$ for strong induction. Example from quora:
Let’s say you wanted to prove that every positive integer has a prime factorization $$𝑝_1𝑝_2𝑝_3...𝑝_𝑚$$.
Let 𝑃(𝑛) be the statement that an integer 𝑛 has a prime factorization. We’ll proceed by strong induction. The basis is pretty clear, so I’ll leave it out.
Next we’ll assume that 𝑃(1),𝑃(2),𝑃(3),...,𝑃(𝑘) are true. 𝑘+1 can either be prime or composite, and if it’s prime we’re done, so we’ll assume it’s composite. That means 𝑘+1 can be written as a product of two positive integers, i.e. 𝑘+1=𝑝𝑞, with $$𝑝,𝑞∈ℤ^+$$. We can write 1<𝑝<𝑘+1, and 1<𝑞<𝑘+1, which implies that 2≤𝑝≤𝑘 and 2≤𝑞≤𝑘.
Here is why we need strong induction: if we had simply supposed 𝑃(𝑛) was true for arbitrary 𝑛, we would be stuck. However, we supposed that 𝑃(𝑛) was true for every positive integer up to 𝑛=𝑘, so we have much more information to work with. Because we supposed this, we know that 𝑃(𝑝) and 𝑃(𝑞) are true, i.e. that 𝑝 and 𝑞 can be represented as a product of primes. We were able to reduce the problem down to a point where 𝑝 and 𝑞 were in a range, and since our inductive hypothesis in strong induction supposes that 𝑃(𝑛) is true for a range of values (rather than just one arbitrary 𝑛), we can now use it to prove the truth of 𝑃(𝑘+1).
Using ordinary induction, I'd say that $$P(p)$$ and $$P(q)$$ are true because $$2≤𝑝≤𝑘$$ and $$2≤𝑞≤𝑘$$ and $$P(k), k\geq 1$$. Why can I not use ordinary induction here? | {
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Another example is the proof that McCarthy 91 function equals 91 for all positive integers less than or equal to 101. The property is $$P(n)=M(101-n), n \geq 0$$ and $$M(n)$$ is the McCarthy function. The author of the proof calculates the base case for $$P(0)$$, then does a supposition that $$P(i), 0 \leq i \leq k, k \geq 0$$. The use of strong induction is justified by the fact that we need the inductive hypothesis to hold for $$k-10$$, but I don't see why $$P(n), n\geq0$$ wouldn't hold for $$n=k-10, k\geq11$$, that is $$n$$ is at least 1, if ordinary induction was used.
• For ordinary induction, you wouldn't assume $P(p)$ and $P(q)$. When proving the theorem for $k+1$, you would assume $P(k)$, and only $P(k)$ (otherwise you are doing strong induction again). And that doesn't help you at all for the prime numbers. – Dirk Jan 27 '20 at 11:40
• @Dirk Technically, you can do it with weak induction. Weak and strong are probably equivalent. However, in this case, using weak looks like a complete mess. – Arthur Jan 27 '20 at 11:45
• @Dirk right, in weak induction I make an assumption that the property holds only for the term immediately before (k+1) and prove that it holds for (k+1). p and q from the proof above do not necessary equal k, therefore I can't assume that P(q) and P(p) are true, and that's where strong induction helps with its "extended" assumption. Is it correct? – super.t Jan 27 '20 at 14:25 | {
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Ordinary or weak induction proves $$Q(n)$$ for all $$n\ge1$$ with a base step $$n=1$$ and an inductive step from $$n=k$$ to $$n=k+1$$.
Complete or strong induction considers the special case where $$Q(n)$$ denotes "$$P(k)$$ for all $$k$$ from $$1$$ to $$n-1$$ inclusive". If we try to prove $$Q(n)$$ for all $$n\ge1$$ by weak induction, the base step is vacuously true, and the inductive step is showing that, if $$P(k)$$ for all $$k$$ from $$1$$ to $$n-1$$ inclusive, then $$P(n)$$. If we can prove this statement, the weak induction on $$Q$$ succeeds, and we have also proven $$P(n)$$ for all $$n\ge1$$.
In other words, strong induction states this: if "$$P(k)$$ for all $$k$$ from $$1$$ to $$n-1$$" implies $$P(n)$$, then $$P(n)$$ for all $$n\ge1$$. Usually, the $$n-1$$ is called $$n$$ instead, so we need to prove "$$P(k)$$ for all $$k$$ from $$1$$ to $$n$$" implies $$P(n+1)$$.
Unlike weak induction, strong induction does not in general need a base step. However, in some cases the argument proving its inductive step has to consider small values of $$n$$ as special cases. | {
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# Surjections using Stirling-numbers
I have a task in which I can't really see what I'm doing wrong. The recommended answer differs from mine and it kind of makes sense to me, but I can't see what's wrong with my solution. The problem is as follows;
In a family of 4 all the 7 deadly sins are present. Every family member exercises(?)
atleast 1 of the 7 sins, but there are no sins that two distinct family members 'exercise'.
(The original problem formulation is not in english, sorry)
The task is to find out how many combinations there are if no single family member can exercise both greed and gluttony.
Basically what I did was to find out the total amount of combinations without taking the constraint into consideration (Surjection from a 7-subset to a 4-subset; $4! * S(7,4)$. After that I wanted to subtract the 'illegal' combinations. The way I figured I would find them out is as follows:
There are 4 possible family members that can exercise both greed and gluttony. After that we have 5 additional sins to 'hand out' amongst all the remaining people (including the one that we already 'handed' both gluttony and greed) so I figure'd we'd then get the total amount of 'illegal' combinations as $4 * 4! * S(5,4)$ but in the recommended solution the amount of illegal combinations was $4! * S(6,4)$ - which I understand (I guess you 'group up' both gluttony and greed as one sin and count the surjections).
But I don't understand why my solution is wrong. Any help greatly appreciated, sorry for the wall of text! | {
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• I don't think so ("But there are no sins that two distinc family members 'exercise') - shouldn't that point to the contrary? Maybe I translated it bad but it says in the text that one specific sin can't be 'exercised' by two family members :) – Nyfiken Gul Oct 18 '16 at 14:36
• Oops; wasn’t thinking clearly. But I do now see the problem: $S(5,4)$ counts only the partitions of the other five sins into four non-empty pieces, so your calculation misses the combinations in which one person has greed and gluttony and nothing else. – Brian M. Scott Oct 18 '16 at 14:38
• Aha! That actually makes perfect sense, didn't consider that at all. Thanks again Brian! ;-) I'll gladly mark it answered if you post your comment as an answer, you're the man! – Nyfiken Gul Oct 18 '16 at 14:44
The problem here is that $S(5,4)$ counts only partitions of the other $5$ sins into $4$ non-empty parts; when you then distribute the greed/gluttony pair to one of those parts, you ensure that the family member who gets that pair also gets at least one other sin. Thus, you’re missing all of the arrangements in which one family member gets greed and gluttony and nothing else. | {
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# Finding the gradient of the restricted function in terms of the gradient of the original function
The following question showed up as part of a proof that I am doing for my research thesis.
If we have a differentiable function $$f: \mathbb{R}^n \to \mathbb{R}$$ and then set $$n-d$$ coordinates to zero we get a new differentiable function $$g: \mathbb{R}^d \to \mathbb{R}$$. Now, given the gradient $$\nabla_x f(x)$$, how one can get $$\nabla_y g(y)$$?
My try
Let $$x \in \mathbb{R}^n$$ and $$S \subset \{1,\dots,n\}$$ such that $$|S|=d$$ where $$|\cdot|$$ is the cardinality of the set. Let $$U_S$$ be a restricted identity matrix such that the $$j$$-th entry of the diagonal matrix is maintained if $$j \in S$$ otherwise it is set to zero. Also, let $$I_S$$ be the restriction of $$U_S$$ where we keep nonzero columns and remove zero columns. Hence,
$$g(y)=f(U_Sx)$$ where $$y=I_S^{\top}x$$.
The above is the translation of what I stated in terms of functions $$f$$ and $$g$$.
From this point things are a little bit unclear. I think the answer should be $$\nabla_y g(y)=I_S^{\top} \nabla_x f(x)$$ but I do not know how to get it.
Also, I know using the chain rule $$J_x f(U_S x)=J_{W} f(W)J_x W= J_{W} f(W)U_S$$ where $$J$$ is the Jacobian and $$W=U_S x$$. In addition, $$\nabla^{\top}_x f(U_Sx) = J_x f(U_S x)=J_{W} f(W)U_S$$. I do not know how to put things together.
Since no one has posted an answer yet, and I get the same result as you suggest, I thought I'll post my solution for you to judge: | {
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We have that $$U_S x = I_S y$$ so that, vieweing matrices as linear transformations $$g(y) = f(U_S x) = f(I_S y) = f\circ I_S (y)$$ And similar to what you write about $$J_{U_S}(x)$$ we have $$J_{I_S}(y) = I_S$$. Applying the chain rule: $$J_{h_1 \circ h_2}(a) = J_{h_1}(h_2(a))J_{h_2}(a)$$ then gives \begin{align} (\nabla_y g(y))^T = J_g(y) =\\ J_{f\circ I_S}(y) = \\ J_f(I_s y)J_{I_S}(y) = \\ J_f(U_s x)I_S = \\ (\nabla_x f(U_S x))^TI_S \implies \\ \nabla_y g(y) = [(\nabla_x f(U_S x))^TI_S]^T = I_S^T\nabla_x f(U_S x) \end{align}
Due to the definitions of $$U_S$$ and $$I_S$$, the zero columns in $$I_S^T$$ exactly matches the rows where $$\nabla_x f(U_S x)$$ and $$\nabla_x f(x)$$ might differ, so finally we obtain $$\nabla_y g(y) = I_S^T\nabla_x f(U_S x) = I_S^T \nabla_x f(x)$$
Edit:
As pointed out in the comments, it would be more correct to write $$\nabla_y g(y) = I_S^T\nabla_x f(I_S y)$$
• Could you make it a little more clear what's going on? Sep 17, 2021 at 10:49
• @Mathemagician314 My answer was a bit confused, I at least had some unnecessary steps. Are there any steps in particular you find strange? Sep 17, 2021 at 11:04
• @Paradox: it is not correct since $U_S\in \mathbb{R}^{n \times n}$ so $U_Sx$ is a vector in $\mathbb{R}^n$ and $I_sx \in \mathbb{R}^{d}$ Sep 18, 2021 at 16:48
• @Sepide I assume you mean $I_Sy \in \mathbb{R}^d$ since I've not written $I_S x$ anywhere. As far as I can see, $I_s$ is an $n \times d$ matrix, not $d \times n$, since it was the non-zero columns that was removed from $U_S$, not non-zero rows. So $I_s y \in \mathbb{R}^n$. Sep 18, 2021 at 16:56
Let fat matrix $${\bf S} \in \Bbb R^{d \times n}$$ be
$${\bf S} := \begin{bmatrix} {\bf I}_d & {\bf O} \end{bmatrix} {\bf P}$$
where $${\bf P}$$ is an $$n \times n$$ permutation matrix. Note that
$${\bf S} {\bf S}^\top = \begin{bmatrix} {\bf I}_d & {\bf O} \end{bmatrix} \underbrace{\,{\bf P} {\bf P}^\top}_{= {\bf I}_n} \begin{bmatrix} {\bf I}_d \\ {\bf O} \end{bmatrix} = {\bf I}_d$$ | {
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Let vector fields $$\rho : \Bbb R^n \to \Bbb R^d$$ and $$\eta : \Bbb R^d \to \Bbb R^n$$ be defined by
$$\rho := ({\bf x} \mapsto {\bf S} {\bf x}), \qquad \eta := ({\bf y} \mapsto {\bf S}^\top {\bf y})$$
and note that $$\rho \circ \eta = \mbox{id}_{\Bbb R^d}$$. Colloquially, if one "expands" and then "restricts", one ends up exactly where one started.
Given differentiable scalar field $$f : \Bbb R^n \to \Bbb R$$, let scalar field $$g : \Bbb R^d \to \Bbb R$$ be defined by
$$g := f \circ \eta$$
Hence,
\begin{aligned} g \left( {\bf y} + {\rm d} {\bf y} \right) = f \left( {\bf S}^\top {\bf y} + {\bf S}^\top {\rm d} {\bf y} \right) &= f \left( {\bf S}^\top {\bf y} \right) + \left\langle \nabla f \left( {\bf S}^\top {\bf y} \right) , {\bf S}^\top {\rm d} {\bf y} \right\rangle \\ &= f \left( {\bf S}^\top {\bf y} \right) + \left\langle {\bf S} \, \nabla f \left( {\bf S}^\top {\bf y} \right) , {\rm d} {\bf y} \right\rangle \end{aligned}
and, thus, the gradient of $$g$$ is
$$\nabla g \left( {\bf y} \right) = \color{blue}{{\bf S} \, \nabla f \left( {\bf S}^\top {\bf y} \right)}$$
or, more succinctly,
$$\boxed{ \qquad \\ \qquad \nabla g = \color{blue}{\rho \circ \nabla f \circ \eta \qquad \\ \qquad}}$$
$$\def\p{\partial} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}}$$For typing convenience, name the gradient \eqalign{ p = \grad{f}{x} \\ } and rename the matrices $$I_S\to S$$ and $$U_S\to U$$.
Also note that $$\,U=SS^T\,$$ and that $$y=S^Tx \qiq dy=S^Tdx$$ Write the differential of the function and rearrange it to recover the desired gradient. \eqalign{ g(y) &= f(Ux) \\ dg &= df \\ &= p:d(Ux) \\ &= p:\LR{U\,dx} \\ &= p:\LR{SS^T\,dx} \\ &= \LR{S^Tp}:\LR{S^T\,dx} \\ &= \LR{S^Tp}:dy \\ \grad{g}{y} &= S^Tp \\ } | {
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In the preceding, a colon is used to denote the Frobenius product, which is a concise notation for the trace \eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\ A:A &= \big\|A\big\|^2_F \\ } The properties of the underlying trace function allow the terms in a Frobenius product to be rearranged in many different but equivalent ways, e.g. \eqalign{ A:B &= B:A \\ A:B &= A^T:B^T \\ C:AB &= CB^T:A = A^TC:B \\ } | {
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# Stability of Tikhonov Regularization
$$\underset{x\in X}{\arg\inf}\left\{||Ax-b||^2+\lambda ||x||^2\right\}$$
I have read that the solution keeps the residual $$||Ax-b||^2$$ small and is stabilized through the $$\lambda ||x||^2$$ term. Can anyone help me understand why that is? I can see that the term prevents overfitting, but I can't quite see how it helps stabilizing.
Assuming $$\|\cdot\|$$ is the $$L_2$$ norm, the solution for $$x$$ is \begin{align*} x = (A^T A + \lambda I)^{-1}A^T b \end{align*} The instability in this solution lies in the inverse. If $$A$$ have columns which are nearly linearly dependent, then $$A^TA$$ is "nearly non-invertible". In other words, the condition number will be very large. The $$\lambda I$$ helps stabilize this inverse, and will always lower the condition number.
• Yes, it is the $L^{2}$ norm that is used. That makes sense! I completely forgot that the solution was relying on $(A^{T}A+\lambda I)^{-1}$. Thank you for the response! – James Feb 4 at 20:11
In order to understand Thikonov's stabilization, it helps to first look at the ordinary least square solution $$x^*$$:
\begin{align*} x^* = (A^T A)^{-1}A^T b \end{align*}
We see that it's necessary to calculate the inverse of $$A^T A$$ and this might not be possible, if $$A$$ has nearly linearly dependent columns. But let's take a closer look onto this, by factorizing just the suspicious term by a singular value decomposition.
Then $$A^T A = U \Sigma V^T$$ where $$U$$ and $$V$$ are the eigenvectors and $$\Sigma$$ is a diagonal matrix that contains the non-zero eigenvalues. It's not of special interest here that $$U=V$$, but it is very important that the pseudoinverse $$(A^T A)^{-1}$$ is found by inverting $$\Sigma$$. | {
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More specific, the reciprocal of each eigenvalue, let's say $$\sigma_i$$ has to be found. And this might get difficult, if two columns are nearly linear dependent. In this case, $$\sigma_i$$ is very small and the result of the division gets very large and tiny pertubations of $$\sigma_i$$ lead to large fluctuations of the inverse. It is possible to monitor such cases and as this answer already mentions the condition number is one of these indicators.
The solution that Thikonov provides to overcome the problem is simple, but very effective: just take a positive variable $$\lambda$$ and add it to the denominator. This will bound the overall result and stabilize the solution:
$$$$\Sigma_{ii}^+ = 1 / (\sigma_i + \lambda)$$$$
As we now have identified the cause of instabilities and inserted a term that prevents them, we can add the same to our known equation and role it back: $$$$U (\Sigma + \lambda I) V^T = U \Sigma V^T + \lambda U V^T = A^TA + \lambda I$$$$
And finally, we arrive at the well known:
\begin{align*} x^* = (A^T A + \lambda I)^{-1}A^T b \end{align*} | {
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# Definition of Inverse in Linear and Abstract Algebra
In a linear algebra text, the following is the definition of the inverse of a matrix
An $n\times n$ matrix $A$ is invertible when there exists an $n \times n$ matrix $B$ such that $$AB = BA = I_n$$
And likewise in an abstract algebra textbook, the definition of the inverse of a group is
Given that $G$ is a group with operation $*$, for each $a \in G$, there exists an element $a^{-1}$ such that $$a*a^{-1} = a^{-1}*a = e,$$ where $e$ is the identity element in $G$. Such element is called the inverse of $a$ in $G$.
Unfortunately, the second semester of abstract algebra didn't quite finalize due to low enrollment, so I'm doing independent self-study of topics I missed in Linear Algebra (as mind preparation). Here's my question:
Is it sufficient to show that $AB = I_n \;\;\implies \;\;B = A^{-1}$ and $A = B^{-1}$? Or must you check both that $AB = I_n$ and $BA = I_n$ to completely conclude that $A = B^{-1}$ and $B = A^{-1}$?
I remember on an exam, I had to prove that for a group homomorphism $\phi: G\to H$, for any $a \in G$, $\phi(a^{-1}) = [\phi(a)]^{-1}$ which I proved by asking the reader to observe that $\phi(a)\phi(a^{-1}) = \phi(aa^{-1}) = \phi(e_G) = e_H$ which because $\phi(a)\phi(a^{-1}) = e_H$, this can only mean that $\phi(a)^{-1} = \phi(a^{-1})$ by definition of inverse. And I got full points for it, but it leaves me wondering: am I supposed to check both arrangements to create the strongest possible argument? | {
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• math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – mvw Feb 8 '16 at 4:52
• For the second problem concerning the homomorphism, you already know that the inverse exits because you are in a group. Proving $\phi (a^{-1})=[\phi(a)]^{-1}$ is not equivalent to proving an inverse exists. – Oliver Jones Feb 8 '16 at 4:57
• In general, if we are following the equivalent conditions for a nonsingular matrix, since $A$ is nonsingular, we know that $A^{-1}$ exists. I suppose a better way to ask my question is: Suppose we have two elements $A$,$B$ and $A$ has an inverse; does it suffice to say that $B$ is the inverse of $A$ if we just show one arrangement (i.e., $AB = e$) gives the identity? So what it seems from other answers, the left inverse and right inverse might not be the same which is why you must check both that $AB = e$ and $BA = e$, no? – Decaf-Math Feb 8 '16 at 5:02
If you already know that $G$ is a group, then to prove that $a$ and $b$ are inverses, it is enough to check $ab = e$.
If $G$ is not a group or you don't yet know that $G$ is a group (for example, if you are trying to prove that $G$ is a group), then it is not enough to show $ab =e$. You also need to show that $ba = e$.
For matrices over a field, $AB = I$ automatically implies that $BA = I$ if $A$ and $B$ are square. Otherwise, if $A$ is $m \times n$ and $B$ is $n \times m$, where $m < n$, then $BA = I$ is impossible. This can be shown by an argument using ranks.
For square matrices over many rings, it is indeed sufficient to check that $AB=I$ to conclude $A=B^{-1}$. It is certainly true for matrices over fields like the real or complex numbers. You can read about it in the link If $AB = I$ then $BA = I$
The class of rings for which this works is called the class of stably finite rings. It is a very broad class of rings, and my guess is that you are working with such a ring. | {
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But in the broadest generality, if you are working in some wild monoid (like a maid ring over a non-stably finite ring) the. It comes necessary to check both $ab$ and $ba$.
In any group (including groups of invertible matrices) it is sufficient to check that something is either a left inverse or a right inverse. This is because in any group, the inverse necessarily exists (by definition of something being a group) and is unique (since $ag=e$ implies $g=a^{-1}e=a^{-1}$ by multiplying on the left by $a^{-1}$, and this implies $ga=e$ by multiplying on the right by $a$).
However, in an arbitrary algebraic structure, a right inverse and a left inverse may not agree.
To show an inverse for both a (square!) matrix and a group element, you must show that it is both a left and a right inverse. Matrix multiplication and general group operations are not commutative.
For homomorphisms sending inverses to inverses, you have more information, which you employed. | {
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# basic algebra question
Albert repays three installments, a, b and c, of a loan that he had taken for buying an electric heater. The total of the first installment and the second installment is 150. The total of the second installment and the third installment is 200. The total of the third installment and thrice the first installment is 250. What is the third installment?
it was asked in my interview for clerical work.
-
Can you translate the given information into (three linear) equations? Could you then solve the system of equations? – David Mitra Jul 6 '12 at 15:39
So you have the system
\begin{align} a + b &= 150 \\ b + c &= 200 \\ 3a + c &= 250 \end{align}
And then you must solve it. Do you know how?
One way might be to subtract the second from the first to get $a + b - (b + c) = a - c = 150 - 200 = -50$, and then to add that to the third to get $a-c + (3a + c) = 4a= -50 + 250 = 200$. This says that $4a = 200$, so that $a = 50$. Then $b = 100$ and $c = 100$.
-
how did u decide that one should substract equation 2 from 1 ... – user286035 Jul 6 '12 at 15:49
@user286035 Because doing so eliminates the variable $c$ – ItsNotObvious Jul 6 '12 at 15:59
Systematic elimination is usually the best way to go. Sometimes there are reasonable alternatives. For example, "add" all three equations. We get $4a+2b+2c=600$. But $2b+2c=400$, so $4a=200$, and therefore $a=50$.
Now easily we see that $b=100$ and therefore $c=100$.
Remark: Or perhaps the mental arithmetic is easier if from $4a+2b+2c=600$ we conclude that $2a+b+c=300$, so $2a=100$, and therefore $a=50$.
-
i liked your way. The mental arith hint is very cool. But when to use which technique ? – user286035 Jul 8 '12 at 7:44
@user286035: For "real world" equations with usually messy coefficients, use systematic (Gaussian) elimination. In interview situations, hard to know, there likely is a trick. Exploit any symmetry. – André Nicolas Jul 8 '12 at 13:02 | {
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# Find the sum of all the multiples of 3 or 5 below 1000
How to solve this problem, I can not figure it out:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
• @J.M.:Much better title thank you! – AD. Nov 7 '10 at 13:42
• @AD: Sometimes straightforward is beautiful. ;) – J. M. is a poor mathematician Nov 7 '10 at 13:44
• This is actually Project Euler problem no 1 and can be solve efficiently by using mutual inclusion exclusion. – Quixotic Nov 7 '10 at 13:45
• This is the postage stamp lemma! Every number greater than 7 can be expressed as $3x+5y$ with $x,y>0$ crazyproject.wordpress.com/2010/10/22/… – N8tron Jun 28 '12 at 13:10
The previously posted answer isn't correct. The statement of the problem is to sum the multiples of 3 and 5 below 1000, not up to and equal 1000. The correct answer is \begin{eqnarray} \sum_{k_{1} = 1}^{333} 3k_{1} + \sum_{k_{2} = 1}^{199} 5 k_{2} - \sum_{k_{3} =1}^{66} 15 k_{3} = 166833 + 99500 - 33165 = 233168, \end{eqnarray} where we have the used the identity \begin{eqnarray} \sum_{k = 1}^{n} k = \tfrac{1}{2} n(n+1). \end{eqnarray} | {
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• The one who posted the answer 233168, please explain that answer in detail . It would be really helpful for us if you will explain. – user126936 Feb 7 '14 at 9:16
• The first two sums account for multiples of $3$ and $5$, the last sum accounts for over-counting multiples of $15$ (which can appear in either of the first two sums). – user02138 Feb 7 '14 at 13:12
• For n=10; and k =3; ( 3 * (3+1) / 2) * 3 – shiva Jul 8 '14 at 10:02
• how did you know to do that? – zurbergram Jun 30 '15 at 3:19
• There is a general principal in counting things in several sets called "inclusion-exclusion". If the things can be in any of several sets, add the totals for each set, subtract the number of things that are in exactly two of the sets, add the number of things that are in exactly three of the sets, etc. – richard1941 Sep 12 '18 at 6:10
First of all, stop thinking on the number $1000$ and turn your attention to the number $990$ instead. If you solve the problem for $990$ you just have to add $993, 995, 996$ & $999$ to it for the final answer. This sum is $(a)=3983$
Count all the #s divisible by $3$: From $3$... to $990$ there are $330$ terms. The sum is $330(990+3)/2$, so $(b)=163845$
Count all the #s divisible by $5$: From $5$... to $990$ there are $198$ terms. The sum is $198(990+5)/2$, so $(c)=98505$
Now, the GCD (greatest common divisor) of $3$ & $5$ is $1$, so the LCM (least common multiple) should be $3\times 5 = 15$.
This means every number that divides by $15$ was counted twice, and it should be done only once. Because of this, you have an extra set of numbers started with $15$ all the way to $990$ that has to be removed from (b)&(c).
Then, from $15$... to $990$ there are $66$ terms and their sum is $66(990+15)/2$, so $(d)=33165$
The answer for the problem is: $(a)+(b)+(c)-(d) = 233168$
Simple but very fun problem. | {
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The answer for the problem is: $(a)+(b)+(c)-(d) = 233168$
Simple but very fun problem.
• +1 but would be a bit better with an explanation as to why one should focus on 990 instead of 1000. – jmoreno Nov 13 '16 at 7:04
• Think this answer is copied from Project Eulers' answer by Rudy: projecteuler.net/thread=1#209 – Eugene Kulabuhov Nov 25 '16 at 21:59
The multiples of 3 are 3,6,9,12,15,18,21,24,27,30,....
The multiples of 5 are 5,10,15,20,25,30,35,40,45,....
The intersection of these two sequences is 15,30,45,...
The sum of the first numbers 1+2+3+4+...+n is n(n+1)/2.
The sum of the first few multiples of k, say k+2k+3k+4k+...+nk must be kn(n+1)/2.
Now you can just put these ingredients together to solve the problem.
To find n use 1000/3 = 333 + remainder, 1000/5 = 200 + remainder, 1000/15 = 66 + remainder and then sum multiples of 3: $3\cdot 333(333+1)/2 = 166833$. multiples of 5: $5\cdot 200(200+1)/2 = 100500$ and subtract multiples of 15 $15\cdot 66(66+1)/2 = 33165$ to get 234168.
• The answer that you have provided is incorrect. Take a look at the accepted answer and see how you can improve your own post. – Jeel Shah Jul 11 '13 at 23:59
• Why can't you duplicate the intersections? – JohnOsborne Jul 29 '18 at 22:32
Well the main equation is already given above. The only question which give me trouble is that why I have to subtract the sum of 15?! Well, the answer is, 15 can be evenly divide by both 3 & 5. So the products of 15 can also be divided by those number as well! So, when you adding the numbers with Sum Of Three & Sum Of Five there are some numbers(i.e. 15,30,45,60....) which are available at both SUMMATION. So, you have to subtract at least once from the total sum to get the answer!
Hope this helps someone like me:) !!
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Hope this helps someone like me:) !!
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# Global extrema question
1. Mar 11, 2005
### mathrocks
I'm supposed to find the absolute max and min for f(x,y)=sin(x) +cos(y) on the rectangle defined by 0<=x<=2*pi 0<=y<=2*pi. Can someone point me in the right direction. I got the critical points but i dont know what to do after that.
2. Mar 11, 2005
### Jimmy Snyder
This can be solved with a few elementary facts concerning the sin and cos functions and no calculus at all.
First, disregarding the rectangle for a moment, it is obvious that f(x,y) must lie between -2 and 2 inclusive for any x and y, because both sin(x) and cos(y) lie between -1 and 1 inclusive, and f(x,y) is the sum of these two.
Second, it is easy to find a pair (x,y) in the rectangle where the values of -2 and 2 are actually attained. For instance if x = PI/2 and y = 0 then f(x,y) = 2. I leave -2 for you to do.
Since f(x,y) attains the values of -2 and 2 on the rectangle, and these values cannot be exceeded for any (x,y), they must be the absolute max and min.
3. Mar 11, 2005
### mathrocks
How do I do this using calculus though, so I know what to do for other problems that arise. I know I must find the partial derviative for f(x,y) and find the critical values of each of those, which I believe are (pi/2, pi), (3*pi/2, 2*pi). Then I'm suppose to do something with the rectangle but I'm not sure what?
4. Mar 11, 2005
### Jimmy Snyder
You need to check the value of f(x,y) along the perimeter of the rectangle.
Consider the fact that the maximum value of the function f(x) = x on the interval from 0 to 1 occurs at the point x = 1 where the maximum value is f(1) = 1. At that point, f'(x) = 1, not 0. So looking for points where f'(x) = 0, while necessary, is not sufficient. Loosely speaking, you need to examine the edges of the domain as well as the points where the derivative is zero.
5. Mar 11, 2005
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5. Mar 11, 2005
### mathrocks
Ok, I'm starting to understand this a little better now. Now my new question is when you have the domain be something like 0<=x<=2*pi, 0<=y<=2*pi, do you have to actually go through each value and see which combination gives you a maximum? Because in the case of f(x,y)=sin(x)+cos(y), the endpoints dont yield a maximum.
6. Mar 11, 2005
### Jimmy Snyder
In general, yes. The reason is somewhat technical, and if you don't understand the rest of this paragraph, you can just accept the answer and safely skip on to the next one. Loosely speaking again, you cannot be assured that the 'edge' of the domain allows analysis. For instance, let the domain be the interior of the rectangle along with some points along the edge thrown in for good measure. Then finding the max of a function on that domain could require you to evaluate the function at those extra points on the edge one by one. Of course, if there were a lot of them, then you could be in it for the long haul. The good news is that the teacher would have the same amount of work to do in order to check your answer, so it probably won't happen to you while you are in school.
However in particular cases you may be able to do better. In fact, for the case at hand, you can look for the extrema along each of the 4 edges by looking for points where the partials are 0 and then evaluate the function at each of the 4 corners.
7. Mar 11, 2005
### mathrocks
That's the thing, when I look for the points where the partials are 0, I get
(x=pi/2), (x=3*pi/2) and (y=pi), (y=2*pi). And when I evaluate the function at the corners which are (x=0), (x=2*pi), (y=0), (y=2*pi), I get x=0, x=0, y=1, y=1. So I dont see how I can get -2,2 even though I know I should.
8. Mar 11, 2005
### Jimmy Snyder
So you need to look at the following points:
1. The points in the interior of the rectangle where the partials are 0.
2. The points along the sides where the partials are 0.
3. The corners. | {
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2. The points along the sides where the partials are 0.
3. The corners.
The largest value of f(x,y) among the union of these three sets of points is the absolute maximum. Similarly for the minimum value.
9. Mar 11, 2005
### dextercioby
$$f(x)=:\sin x+\cos y$$
The condition for extremum-------->critical points.
$$\frac{\partial f}{\partial x} =0 ;\frac{\partial f}{\partial y}= 0$$
The type of extremum---------->
Hess(f)|_{sol.critical points} ? 0
Daniel.
P.S.If the hessian is the positive,then it's a minimum,if it's negative,it's a maximum,if it's zero,it's a saddle point... | {
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# Linear Algebra, cube & dimensions > 3
I have found interesting problem in Gilbert's Strang book, ,,Introduction to Linear Algebra'' (3rd edition):
How many corners does a cube have in 4 dimensions? How many faces? How many edges? A typical corner is $(0,0,1,0)$
I have found the answer for corners:
We know, that corner is $(x_1,x_2,x_3,x_4)$. For every $x_i$ we can use either $1$ or $0$. We can do this in $2 \cdot 2 \cdot 2 \cdot 2 = 2^4 = 16$ ways.
The same method can be used for general problem of cube in $n$ dimensions (I suppose):
Let's say, we have $n$-dimensional cube (I assume, that length of edge is $1$, but it can be some $a$, where $a \in \mathbb{R}$ [1]). Here, corner of this cube looks like this: $(x_1,x_2, \ldots , x_n)$. For every $x_i$ there are $2$ possibilities: $x_i = 0$ or $x_i = 1$ ($x_i = a$ in general). So, this cube has $2^n$ corners.
It was pretty simple, I think. But now, there are also faces and edges. To be honest, I do not know, how to find the answer in this cases.
I know, that solution for this problem is:
A four-dimensional cube has $2^4 = 16$ corners and $2 \cdot 4 = 8$ three-dimensional sides and $24$ two-dimensional faces and $32$ on-dimensional edges.
Could You somehow explain me, how to figure out this solution? I have found solution for corners by myself, using Linear Algebra methods & language. Could You show me, how to find the number of edges and faces, using Linear Algebra methods?
Is there other method to find these numbers? (I suppose, that answer for this question is positive)
I am also interested in articles/textbooks/etc. about space dimensions, if You know some interesting positions about that, share with me (and community).
As I wrote: I am interested in mathematical explanations (in particular using Linear Algebra methods/language but other methods may be also interesting) and some intuitions (how to find solution using imagination etc. [2]).
Thank You for help. | {
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Thank You for help.
[1] I am not sure of this assumption, because:
(a) I am not sure, how edges (and faces) behave in $n$ dimensions
(b) I am not sure, how should I think about the distance in $n$ dimensions. I mean, I know, that my intuition may play tricks here
[2] I am not asking, how to imagine $4$ dimensional cube, but I think, that there is a way to find the solution, using reasoning, not only Linear Algebra.
My definition of face (there was a comment about that) is the same as definition here: http://en.wikipedia.org/wiki/Face_(geometry), especially:
In geometry, a face of a polyhedron is any of the polygons that make up its boundaries.
• You are right that cubes have the same shape (number of faces) regardless of the side length. The usual distance is $\sqrt{(x_1-y_1)^2+(x_2-y_2)^2 \ldots +(x_n-y_n)^2}$ The article that Joseph Malkevitch cites is a good combinatorial approach. – Ross Millikan Jan 6 '11 at 3:10
• @Ross Millikan: I know, that using this formula I can find length in $n$ dimensions. But I can't figure out, how to imagine this, like 1,2,3 dimensions... – exTyn Jan 6 '11 at 15:30
• Gilbert Strang has his lectures on line for free at: ocw.mit.edu/courses/mathematics/… – Tpofofn Jan 7 '11 at 3:26
As you mentioned, a vertex (or a 0-face) is just a choice of string $(x_1,\ldots, x_n)$, where $x_i\in \{0,1\}$. If you think back to dimensions 2 or 3 (or even 1!), you'll see that an edge (i.e., a 1-face) is determined by two vertices which differ in a single slot. So e.g. on the unit square, the left edge goes between $(0,0)$ and $(0,1)$ -- we may name this edge more succinctly as $(0,*)$. This method generalizes, too: the bottom square (2-face) of the 3-cube is $(*,*,0)$, etc. Note that the $n$-cube will have $k$-faces for $0\leq k\leq n$. See if you can find a formula for how many there are! | {
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This isn't exactly linear algebra, it's more like combinatorics. Also, this problem is actually a decent introduction to the idea of higher dimensions, if you try to visualize things in 4 dimensions. Search the internet for some representations of 4-cubes, and try to understand why they look the way they do.
To calculate the number of edges: as you say there are $2^n$ corners. Each one is connected to n other corners. Adding all of these up counts each edge twice, so there are $2^{n-1}n$ edges, which equals $32$ for $n=4$. Another way to count edges is to define $E(n)$ as the number of edges in $n$ dimensions. If you think of the $n+1$ dimensional cube as connecting the corresponding corners of two $n$ dimensional cubes, the recurrence is $E(n+1)=2E(n)+2^n$
Square faces are made by starting with a square in the one of the two $n$ cubes plus the translation of the edges of the $n$ cube in the new dimension. So $S(n+1)=2S(n)+E(n)$.
$3$-cube faces are made by starting with a $3$-cube in one of the two n-cubes plus translation of the squares in the new dimension. So $C(n+1)=2C(n)+S(n)$.
• An expository article about how to think about n-cubes from a combinatorial point of view can be found here: york.cuny.edu/~malk/tidbits/n-cube-tidbit.html – Joseph Malkevitch Jan 5 '11 at 1:23
• @Joseph: Thanks. It shows that the recurrence will be the same for all dimensions of boundary: points, lines, squares, etc. – Ross Millikan Jan 5 '11 at 3:36
First you need to be clear about what you mean by face, edge, vertex, etc. In my view if we have an $n$ dimensional cube, then $n-1$ dimensional cubes form its faces, and $n-2$ dimensional cubes form its edges. Vertices as you define above are 0-D points. We do not have names for $n-3$, $n-4$ dimensional boundaries so I will not attempt to name them here. | {
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If you are looking for an intuitive way to build up the relations you can define inductive relationships using the method of extrusion. You can start with $n=1$ if you like. In this case we have one 1-D cube (i.e. a segment) with two 0-D faces (i.e. points). So
$$C_1 = 1$$ $$F_1 = 2$$ $$E_1 = 0$$ $$V_1 = 2$$
Where $C_1$ is the number of cubes in 1-D space, $F_1$ is the number of faces, $E_1$ is the number of edges (not defined here), and $V_1$ is the number of vertices.
We create a 2-D cube (i.e. a square) by extruding this 1-D cube in a direction orthogonal all current dimensions. By doing this two things happen:
1) Every object is copied maintaining its current dimensions. We had one segment, now we have two, we had two vertices, we now have four.
2) Every object extruded into the new direction creates one object of one more dimension. The 1-D segment is extruded over a second dimension to form a square (i.e. 2-D object). The vertices (0-D) are extruded to become lines.
So the number of cubes, faces, edges, vertices becomes
$$C_2 = 1$$ $$F_2 = 2C_1 + E_1 = 4$$ $$V_2 = 2V_1 = 4$$ $$E_2 = V_2$$
Repeat the process for a 3-D cube. Again we have two sources for new objects, copying and extrusion.
$$C_3 = 1$$ $$F_3 = 2C_2 + E_2 = 6$$ $$E_3 = 2F_2 + V_2 = 12$$ $$V_3 = 2V_2$$
Continue the process to get to higher dimensions.
This is the 27th problem in Exercises 1.1 of the book, Introduction to Linear Algebra, Gilbert Strang. I hope my answer will be helpful for future readers.
Question 1: How many corners does a cube have in 4 dimensions?
Solution: A corner of a cube in $$N$$ dimenison(s) can be written as $$(x_1, x_2, \cdots, x_N)$$ and for each $$x$$, the choice is either $$1$$ or $$0$$. Then the number of corners equal to $$2^N$$. Therefore, the number of corners in 4 dimesions is $$16$$.
Question 2: How many edges does a cube have in 4 dimensions?
Solution:
Notice that there are two useful invariants. | {
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Solution:
Notice that there are two useful invariants.
• Each edge is shared by exactly $$2$$ corners.
• Each corner is connected to $$N$$ corners.
Then the number of edges of a cube in $$N$$ dimensions is $$2^N \cdot N \cdot \frac{1}{2} = 2^{N-1}\cdot N$$. Therefore, the number of edges is $$2^{4-1} \cdot 4 = 32$$.
Question 3: How many 3D faces does a cube have in 4 dimensions?
Solution: Let's first consider the question, "How man 2D faces does a cube have in 3 dimensions?" This answer is $$3 * 2 = 6$$. $$3$$ means 3 dimensions and $$2$$ means positive and negative directions.
Now, by a similar reasoning, the number of 3D faces is $$4*2=8$$.
Question 4: How many 2D faces does a cube have in 4 dimensions?
Solutions:
Notice that there are two useful invariants.
• Starting from a corner, the number of faces between edges is $$C_{N}^{2}$$.
• Each face is shared by $$4$$ corners.
Then the numebr of 2D faces of a cube in $$N$$ dimensions is $$C_{N}^{2} \cdot 2^{N} \cdot \frac{1}{4} = C_{N}^{2} \cdot 2^{N-2}$$. Therefore, the number of 2d faces of a cube in 4 dimensions is $$6*16/4=24$$. | {
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If $X \subseteq A \cup B$, then $X \subseteq A$ or $X \subseteq B$.
If $X \subseteq A \cup B$, then $X \subseteq A$ or $X \subseteq B$.
My counterexample: Let $A = \{1\}$ and $B = \{2\}$. Then $\{1, 2\} \subseteq A\cup B$, but $\{1,2\} \not\subseteq A$ and $\{1,2\} \not\subseteq B$. How would I prove this generally? I've tried starting with the fact that if $X \subseteq A \cup B$ then $x \in X$ implies $x \in A \cup B$ and further $x \in A$ or $x \in B$, but I'm not making any decent progress after that.
How would you prove this without a counterexample?
• You already proved it: the claim is false, and it is enough to give one single counter example...and you did it. – Timbuc Nov 19 '14 at 4:42
• I know that I already proved it. I just want to do try doing it without the counterexample. – St Vincent Nov 19 '14 at 4:44
• To disprove something it suffices to find a counterexample. – Empiricist Nov 19 '14 at 4:46
• @StVincent, you can't. First, because sometimes it is true that $\;X\subset A\;\;or\;\; X\subset B\;$, and second because you must give a counter example to refute a mathematical claim. – Timbuc Nov 19 '14 at 4:47
• I know that the counterexample is sufficient, but I wanted some more practice. In other words, could it be shown false without the counterexample? – St Vincent Nov 19 '14 at 4:48
As my comment has already mentioned, counterexamples are enough to disprove a statement. If you want to characterize when the statement is wrong, an answer goes as follows:
The statement is in general wrong, given that $B \setminus A$ and $A \setminus B$ are non-empty.
Take $X = A \cup B$. Then $X \subseteq A \cup B$ trivially.
Since $B \setminus A$ is non-empty, pick $b \in B \setminus A$. Then $b \in X \setminus A$ and therefore it is false to say $X \subseteq A$. | {
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It is symmetric to show that it is false to say $X \subseteq B$ either. This completes the proof of the statement "Given $B \setminus A$ and $A \setminus B$ are non-empty, $X \subseteq A \cup B$ does NOT imply $X \subseteq A$ or $X \subseteq B$". But please note that we also suggest a counterexample here.
Moreover, this is the largest generality you can get, i.e., the conditions "$B \setminus A$ and $A \setminus B$ are non-empty" cannot be dropped. If, say, $B \setminus A$ is empty, then $B \subseteq A$ and $A \cup B = A$. Then certainly $X \subseteq A \cup B$ implies $X \subseteq A$.
Your counterexample proves that the statement is false.
Counterexamples are perfectly valid counterproofs.
There is nothing else required--you have shown that it is not the case that the claim is true in all cases. | {
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# Math Help - Sum of Probabilities of combinations
1. ## Sum of Probabilities of combinations
Hi,
I'm thinking if you can help me think of a better way to solve this.
Here's the situation
I have probabilities of the number of bags I can sell in a day
2 - 30 %
3 - 28%
4 - 20%
5 - 16%
6 - 6%
Now, I want to know the probability of selling, for example, 8 bags after 3 days.
What I did so far is this,
I get all the permutations of 3 numbers out of {2, 3, 4, 5, 6}, where the sum is 8, so I have the following with their respective probabilities
2,2,4 = .30*.30*.20 = .018
2,3,3 = .30*.28*.28 = .02352
2,4,2 = .30*.20*.30 = .018
3,2,3 = .28*.30*.28 = .02352
3,3,2 = .28*.28*.30 = .02352
4,2,2 = .20*.30*.30 = .018
thus, the probability of selling 8 bags after 3 days is .018+.02352+.018+.02352+.02352+.018 = .12456
My problem is, I need to get all the possible number of bags sold in 3 days with their probabilities of occurring. Is there a way that I can compute for that easily? The true situation is working with more than 5 possible number of bags in a day.
2. ## Re: Sum of Probabilities of combinations
Originally Posted by rikari
Hi,
I'm thinking if you can help me think of a better way to solve this.
Here's the situation
I have probabilities of the number of bags I can sell in a day
2 - 30 %
3 - 28%
4 - 20%
5 - 16%
6 - 6%
Now, I want to know the probability of selling, for example, 8 bags after 3 days.
What I did so far is this,
I get all the permutations of 3 numbers out of {2, 3, 4, 5, 6}, where the sum is 8, so I have the following with their respective probabilities
2,2,4 = .30*.30*.20 = .018
2,3,3 = .30*.28*.28 = .02352
2,4,2 = .30*.20*.30 = .018
3,2,3 = .28*.30*.28 = .02352
3,3,2 = .28*.28*.30 = .02352
4,2,2 = .20*.30*.30 = .018
thus, the probability of selling 8 bags after 3 days is .018+.02352+.018+.02352+.02352+.018 = .12456 | {
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thus, the probability of selling 8 bags after 3 days is .018+.02352+.018+.02352+.02352+.018 = .12456
My problem is, I need to get all the possible number of bags sold in 3 days with their probabilities of occurring. Is there a way that I can compute for that easily? The true situation is working with more than 5 possible number of bags in a day.
It depends on your definition of "easy". The easiest way I know of to solve this problem is to use a generating function. In this case, the generating function is
$f(x) = 0.3 x^2 + 0.28 x^3 + 0.2 x^4 + 0.16 x^5 + 0.06 x^6$
I think you can see how the function is derived from your data. Then if you want to know the probability of selling exactly 8 bags in 3 days, the steps are
1. Expand $(f(x))^3$. This is easy if you have a computer algebra program, or you can just use Wolfram alpha.
2. Extract the coefficient of $x^8$ in the result. You will see in this case, that the coefficient is 0.12456, as you said.
If you would like to see the complete result of the expansion, go to Wolfram alpha
Wolfram|Alpha: Computational Knowledge Engine
and enter
expand (0.3 x^2 + 0.28 x^3 + 0.2 x^4 + 0.16 x^5 + 0.06 x^6)^3
3. ## Re: Sum of Probabilities of combinations
Thanks for the response.
I didn't think of using the generating functions, haven't thought of it yet, but surely a bit easy than what I'm currently doing.
A huge thanks | {
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# Math Help - Finding the slope of a line
1. ## Finding the slope of a line
We just started doing inequalities in class and I get the basics, but this is what gets me.
M = 7, point (0,-5)
How would I write this in Y = mx + b format?
2. $y = m * x + b$
First, we know m:
$y = 7 * x + b$
Next, we know x and y:
$-5 = 7 * 0 + b$
Solve for b:
$-5 = b$
Notice, if $x = 0$, and $(x,y)$ is on the line, $b = y$.
So, now, write your equation down:
$y = 7 * x + -5$
Or:
$y = 7 * x - 5$
3. Originally Posted by largebabies
We just started doing inequalities in class and I get the basics, but this is what gets me.
M = 7, point (0,-5)
How would I write this in Y = mx + b format?
The y intercept = -5, therefore
y=7x-5
4. m=7, (0,-5)
y- (-5)
_______ = 7 : then cross multiply
x-0
y+5= 7x: move five to the other side by subtracting it
so then you get y=7x -5
5. Thank you all, it's help alot
Although, I need help on another equation, what if there is two points?
example: M = 3/2, point (5,-6)?
6. $y = m * x + b$
First, we know m:
$y = \frac 3 2 * x + b$
Next, we know x and y:
$-6 = \frac 3 2 * 5 + b$
Solve for b:
$-6 - \frac {15} 2 = \frac {-27} 2 = b$
So, now, write your equation down:
$y = \frac 3 2 * x + \frac {-27} 2$
7. Thank you once again, although it's supposed to be 27/2 not -27/2
"passes through (4,2) and (0,8)"
9. Originally Posted by largebabies
"passes through (4,2) and (0,8)"
Find the slope:
$m=\frac{y_1-y_2}{x_2-x_1}=\frac{8-2}{0-4}=\frac{6}{-4}=-\frac{3}{2}$
From (0, 8) we know the y-intersept = 8.
Use the slope-intercept form of the general equation:
$y=mx+b$, where m is the slope, and b is the y-intercept.
$y=-\frac{3}{2}x+8$
10. that helped alot!
although i'm now stuck on another equation , I did the same thing with this and got y= -10/-2 when I needed -5/4.
The equation goes like this.
“passes through (-5,10) and (3,0)”
help? | {
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# Formulation and computation of “the” unique median of an even-sized list
Consider an even-sized set of numbers $X = \{x_k\}$, such as $X = \{1, 2, 7, 10\}$.
The median $m$ is defined as:
$$m = \mathrm{arg \min_x} \sum_k \lvert x_k - x\rvert^1$$
Any $m \in [2, 7]$ is a minimizer of this function, and is therefore "a" median of this list.
Now, it is common practice to take the average of 2 and 7 and call it "the" median.
But that's lame, and I think I have invented (?) a more logical way to find a unique median $m^*$:
$$m^* = \lim_{\epsilon \to 0^+} \mathrm{arg \min_x} \sum_k \left\lvert x_k - x\right\rvert^{1+\epsilon}$$
Diferentiation to find the minimum only gets us so far:
$$\sum_k \mathrm{sgn}{\left(x_k - m^*\right)}\left\lvert x_k - m^*\right\rvert^{\epsilon} = 0$$ This expression can be solved numerically for smaller and smaller $\epsilon$ to give $m^* \approx 4.85$ in this example, and I suspect the "correct" median is in fact $m^* = 34/7$, but I don't know how to prove it.
I have 3 questions:
1. First of all, is this a well-known and/or useful approach? Does it have a name?
I came up with the new formulation myself, but I've never seen it used anywhere.
2. Is there some way to directly find the exact value of $m^*$, without numerical optimization?
If not, is there a better/faster approach than brute-force numerical optimization techniques?
3. Is this a (convex?) optimization problem, and if not, can it be reformulated as one?
The trouble here is that I can't find any objective function that has a unique minimum at $m^*$.
The best I can do is to find a generalized function (i.e., the limit of another function), but when I do that, I don't think the problem is a convex optimization problem anymore.
Is there another way to pose the problem that conforms better to existing optimization frameworks? | {
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-
I am confused. If $m^*$ is the minimizer of a function over the elements $x\in X$, then how could $m^*=4.85\notin X$? – NotNotLogical Jun 26 at 20:36
@NotNotLogical: Where did you get the idea that we must have $x \in X$? – Mehrdad Jun 26 at 20:39
Ah, I was misreading. I think I now understand. – NotNotLogical Jun 26 at 20:39
What's the problem with non-uniqueness here? – Peter Sheldrick Jun 26 at 21:48
@PeterSheldrick: Huh? The problem is the very fact that it's not unique, even though there's a perfectly sensible definition that I demonstrated gives a unique answer. It doesn't make sense to say 2 is a median of that set, nor does it make sense to say 7 is, because neither is in the "center" of the data, even though they all minimize the typical objective. Hence it makes sense to look for a better number and thus a better objective. – Mehrdad Jun 26 at 21:52
If the values are sorted $x_1 \le x_2 \ldots \le x_n$, then the value $m^*$ is the unique solution on the interval $[x_{n/2},x_{n/2+1}]$ to the following equation:
$$(m^* - x_1)(m^*-x_2)\ldots(m^*-x_{n/2})=(x_{n/2+1}-m^*)(x_{n/2+2}-m^*)\ldots(x_{n}-m^*).$$
When $n=2$, it's just the mean, and when $n=4$, $m^* = (x_3x_4-x_1x_2)/(x_3+x_4-x_1-x_2)$, which in your example is $(7\cdot 10-1\cdot2)/(7+10-1-2)=34/7$. I don't see any simple way to solve the equation in closed form for higher $n$, other than standard techniques for finding roots of polynomials.
To prove that the above equation defines $m^*$, you just need to go few steps further in the manipulation of the derivative. That is $m^* = \lim_{\epsilon\rightarrow 0^+} m_\epsilon$, where $m_\epsilon$ is the solution to:
$$\sum_{k=1}^n \mathrm{sgn}{\left(x_k - m_\epsilon\right)}\left\lvert x_k - m_\epsilon\right\rvert^{\epsilon} = 0$$
For small $\epsilon$, we should have $x_{n/2}\le m_\epsilon \le x_{n/2+1}$, so this becomes:
$$-\sum_{k=1}^{n/2} (m_\epsilon-x_k)^{\epsilon} + \sum_{k=n/2+1}^{n} (x_k-m_\epsilon)^{\epsilon}=0$$ | {
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$$-\sum_{k=1}^{n/2} (m_\epsilon-x_k)^{\epsilon} + \sum_{k=n/2+1}^{n} (x_k-m_\epsilon)^{\epsilon}=0$$
Expanding to first order in $\epsilon$ gives: $$-\sum_{k=1}^{n/2} (1+\epsilon \log(m_\epsilon-x_k)) + \sum_{k=n/2+1}^{n} (1+\epsilon\log(x_k-m_\epsilon))=O(\epsilon^2)$$
The constant terms cancel, and dividing by $\epsilon$ gives: $$-\sum_{k=1}^{n/2} \log(m_\epsilon-x_k) + \sum_{k=n/2+1}^{n} \log(x_k-m_\epsilon)=O(\epsilon)$$
Then by taking the limit as $\epsilon \rightarrow 0^+$, we get: $$-\sum_{k=1}^{n/2} \log(m^*-x_k) + \sum_{k=n/2+1}^{n} \log(x_k-m^*)=0$$ Which is equivalent to the stated condition.
There are different ways to approximate the 1-norm with a differentiable function, and each approximation will give a different unique "median". I don't know of any reason to prefer any one approximation over another other than convenience.
-
+1 Holy cow, this looks exactly like the kind of answer I was hoping for! So you used approximated $m^* - x_k$ to first-order to be equal to its linearization $1 + \epsilon \log(m^* - x_k)$, because it's equal in the infinitesimal case? It seems so obvious in hindsight but it's very clever, I wouldn't have thought of it for quite a long time! Thanks so much, I learned something new today from your answer. :) – Mehrdad Jun 28 at 0:30
Glad to help. I rewrote the argument to be more rigorous. Hopefully it's clearer now. – p.s. Jun 28 at 1:18
Indeed, very nice! Well done. – Michael Grant Jun 28 at 3:36
I'll bet that this is very amenable to a simple numerical search (Newton and/or bisection) with the midpoint as an initial condition. – Michael Grant Jun 28 at 3:39
It also makes me wonder why first-order approximations are so special. A second-order approximation would seem correct but unhelpful, whereas a zeroth-order approximation would tell us nothing. So what's so special about first-order that gives us the answer we want exactly in the limiting case? Maybe I should ask that as a question... – Mehrdad Jun 28 at 11:54 | {
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Unfortunately, this approach is not compatible with convex optimization in practice.
The reason is that in an optimization context, a convex function and its epigraph are assumed interchangeable. That is to say: consider the following two problems: $$\begin{array}{ll} \text{minimize} & f(x) \end{array}$$ $$\begin{array}{ll} \text{minimize} & y \\ \text{subject to} & f(x) \leq y \end{array}$$ These problems are equivalent if $f$ is convex: that is, given the solution to one, the solution to the other is evident, and vice-versa. Of course, the second one has an associated dual variable while the first one does not, but that doesn't change the equivalence.
Now let's consider your function for $f$, set in the second form above: $$\begin{array}{ll} \text{minimize} & y \\ \text{subject to} & \lim_{\epsilon\rightarrow 0^+} \sum_k | x_k - x |^{1+\epsilon} \leq y \end{array}$$ This is, in all practical respects, equivalent to $$\begin{array}{ll} \text{minimize} & y \\ \text{subject to} & \sum_k | x_k - x | \leq y \end{array}$$ which is of course what you'd get with the standard median function. Any practical system for optimization is really not going to be able to differentiate between the two forms. You could, of course, fix $\epsilon$ to be small and nonzero, but then you've destroyed equivalence, and of course made the leap from a linear problem to a nonlinear one.
Conceptually, what is happening here is that you are preferring a particular element of the arg min set over the rest. But establishing preferences among feasible points is precisely what the purpose of an objective function is. You need to find a way to integrate your preferences more directly into your objective or constraints. For instance, if you are determine to preserve the numerical results induced by the $|\cdot|^{1+\epsilon}$ approach---in particular, if it is important that $34/7$ be the correct answer in this example---then you will not be able to use this median function in convex optimization. | {
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-
Thanks for the response, but I feel it's begging my own question when you say "You need to find a way to integrate your preferences more directly into your objective or constraints."... the entire point of my question was to figure out how I was supposed to do that, because as I had already mentioned, I already realized this was unlikely to work with standard (convex) optimization. Your answer just basically summarized the problem I was facing but didn't help me get anywhere. – Mehrdad Jun 26 at 21:06
I am certainly not claiming to have answered every bullet point. But you did ask if this could be formulated as a (convex?) optimization problem, and I answered it. – Michael Grant Jun 27 at 4:46
Besides: your definition of median is non-standard. You well know that the standard median is unique (4.5). Even if we set aside the midpoint portion of the standard statistical definition, your optimization-based definition includes two numbers (2, 7) that do not satisfy even the fundamental criteria of a median. Your definition is already a convenience, then. Certainly you are not the only one to employ this convenience in practice, but the fact remains. – Michael Grant Jun 27 at 5:08
I did not ask if this could be formulated as a convex optimization problem, I asked if it was a (perhaps convex) optimization problem, and if not, whether it could be re-formulated as such a problem. All you did was tell me, "You need to find a way to integrate your preferences more directly into your objective or constraints". In other words, you just repeated back at me the obvious fact that I need to reformulate it as a convex optimization problem, without helping me actually get anywhere. – Mehrdad Jun 27 at 5:37
I see no reason to accept my answer. It only addresses a part of your question. Nevertheless I will incorporate your edit... – Michael Grant Jun 27 at 22:59 | {
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# Applying method of characteristic equations to $e^y u_x + u_y = u^2$
I have the PDE $e^y u_x + u_y = u^2$, $u(x,0) = x$ for small $|y|$. Also $u = u(x(s), y(s))$.
So $\frac{dx}{ds} = e^{y}$, $\frac{dy}{ds} = 1$, and $\frac{du}{ds} = u^{2}$.
Solving the first ode: $\frac{dx}{ds} = e^{y} \to dx = e^{y}ds$ and integrate both sides to obtain $x = e^{y}s + c_1$.
The second ODE: $\frac{dy}{ds} = 1 \to dy = 1ds$ and integrate both sides to obtain $y = s + c_2$.
The third ODE: $\frac{du}{ds} = u^{2}$, the solution to this ode is $-\frac{1}{u} = s + c_3 \implies u = -\frac{1}{s+c_3}$.
Okay now I want to eliminate the parameter $s$. I solve for $s$ from the solution of the second ode: $s = y -c_2$ then substitute into the solution of the first ode: $x = e^{y}(y -c_2) + c_1 = e^{y}y - e^{y}c_2 + c_1$. Also $u = -\frac{1}{y-c_2 + c_3}$.
This is where I am stuck and am not sure what to do. I was attempting to follow https://en.wikipedia.org/wiki/Method_of_characteristics#Example but I am not doing this correctly I think. | {
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• With the given initial condition, you can show that $c_2 = 0$. Also, you need to substitute $y$ into the characteristic equation $\frac{dx}{ds} = e^y = e^s$ and then integrate; this will give $x = e^s + c_1$ and $c_1$ is found by solving $x(0) = x_0$, where $x_0$ is the parameterisation parameter for the "initial curve". – Chee Han Apr 3 '18 at 17:40
• Why is $e^y = e^{s}$? I know $y = y(s)$. Letting $x(0) = x_0$, from the equation $x(s) = e^s + c_1$, we get $x_0 = e^0 + c_1 = 1 + c_1$. So $c_1 = x_0 - 1$. Is this right? – Taln Apr 3 '18 at 18:04
• Because $y = s + c_2$ and $y(0) = 0$ gives $c_2 = 0$. And yes, $c_1$ is correct. Now you can do the same for $\frac{du}{ds} = u^2$, with $u(0) = x_0$. Then the final step would be to eliminate $s$ and $x_0$ if possible. Depending on the question, you might also want to determine the validity of your solution. Any standard PDE books that discuss first order PDE should have more information about this. – Chee Han Apr 3 '18 at 19:43
• Even with @CheeHan's corrections, it doesn't seem like a solution exists. Are there any typos in the original PDE? – MasterYoda Apr 3 '18 at 20:48
• @MasterYoda I've added $' for small |y| '$ to my post, but other than that I see no typos. – Taln Apr 3 '18 at 20:57
## 1 Answer
$$e^yu_x+u_y=u^2$$ I agree with your equations which can be written in a summarised form : $$\frac{dx}{e^y}=\frac{dy}{1}=\frac{du}{u^2}=ds$$ A first characteristic equation comes from $\quad\frac{dx}{e^y}=\frac{dy}{1}\quad$ leading to : $$e^y-x=c_1$$ A second characteristic equation comes from $\quad\frac{dy}{1}=\frac{du}{u^2}\quad$ leading to : $$\frac{1}{u}+y=c_2$$ The general solution of the PDE, expressed on the form of implicit equation is : $$\Phi\left((e^y-x) \:,\: (\frac{1}{u}+y) \right)=0$$ $\Phi$is an arbitrary function of two variables.
Or, equivalently the general solution of the PDE on explicit form is : $$\frac{1}{u}+y=F(e^y-x)$$ where $F$ is an arbitrary function. $$u(x,y)=\frac{1}{-y+F(e^y-x)}$$ | {
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CONDITION :
$u(x,0)=x=\frac{1}{-0+F(e^0-x)}=\frac{1}{F(1-x)}$
$F(1-x)=\frac{1}{x}$
Let $\quad X=1-x\quad;\quad x=1-X\quad;\quad F(X)=\frac{1}{1-X}$
So, the function $F$ is determined.
We put it into the above general solution where $\quad X=e^y-x\quad$ then $\quad F(e^y-x)=\frac{1}{1-(e^y-x)}$
$$u(x,y)=\frac{1}{-y+\frac{1}{1-e^y+x}}$$ This is the particular solution of the PDE which fits to the boundary condition.
Note: The above calculus is consistent with your calculus. The difference is that $ds$ is eliminated at the beginning, while in your calculus $ds$ is eliminated later. This doesn't change the final result.
ADDITION, answering to the comment of MasterYoda :
Of course, I checked the solution $u(x,y)=\frac{1}{-y+\frac{1}{1-e^y+x}}$ before publishing my answer. After the MasterYoda's comment I checked it again. Definitively the solution satisfies the PDE (copy below).
• Your solution doesn't appear to satisfy the original PDE. – MasterYoda Apr 3 '18 at 21:46
• @ MasterYoda : The solution satisfy the PDE. See the addition to my main answer. – JJacquelin Apr 4 '18 at 7:26
• Thanks, looks like I did my math wrong. – MasterYoda Apr 4 '18 at 23:59 | {
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# Exponential Decay (Precalculus)
I'm working on this problem.
The radioactive element carbon-14 has a half-life of about 5,750 years. The percentage of carbon-14 present in the remains of animal bones can be used to determine age. How old is an animal bone that has lost 40% of its carbon-14?
I'm using the formula $y = Ce^kt$ to model the decay.
To make my life easier, I consider $5,750$ years to be 1 time unit.
So knowing this is a half life, for a single time unit I can say
$$50 = 100 e^k$$
Then
$$\frac{1}{2}=e^k$$
Then log of both sides
$$\ln\frac{1}{2}=\ln e^k$$
$$\ln\frac{1}{2}= k \cdot \ln e$$
So
$$k = \ln\frac{1}{2}$$
Good so far.
Next we need to determine the $t$ after Carbon14 has decayed to $60\%$.
So
$$60 = 100 e^kt$$
which is
$$\frac{3}{5}= e^kt$$
Then I take the log of both sides
$$\ln \frac{3}{5}= \ln e^kt$$
then
$$\ln\frac{3}{5} = kt\cdot \ln e$$
simplifies to
$$\ln\frac{3}{5} = kt$$
Plugging in $k$ we get
$$t =\frac{ \ln\frac{3}{5}}{\ln\frac{1}{2}}$$
Recall that t is a unit where each unit is $5,750$ years, so the final answer is
$$5750\cdot \left(\frac{\ln\frac{3}{5}}{\ln\frac{1}{2}}\right)$$
My answer is $\approx4,237.55$ years.
The answer key says the $4,257$ years. http://www.sosmath.com/cyberexam/precalc/EA3001/EA3001.html
Where did I go wrong?
• It seems to be that site has a mistake. What you did is correct. – DonAntonio Dec 14 '13 at 20:06
• Minor comment: It would be more conventional to use $e^{-kt}$. – André Nicolas Dec 14 '13 at 20:11
## 1 Answer
I personally think you did it right. The structure and execution is correct. I checked my answer over and over again and it is correct. I think the answer key is wrong. | {
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MAT 142 3.6 Conditional Probability and Intersections
# MAT 142 3.6 Conditional Probability and Intersections
398.7k points
1. The Por Mano Tile Company imports handmade tiles from Mexico. Saltillo tiles makes up 67% of their imports and the rest are talavera tiles. Based on past experience, they expect 3.7% of the Saltillo tiles to arrive broken and 5.4% of the talavera tiles to be broken.
a. Make a complete probability tree for this situation.
b. Find the probability that a tile is a talavera tile and is broken
c. Find the probability that a tile is not broken.
d. Find the probability that a Saltillo tile is broken.
e. Find the probability that a talavera tile is not broken
2. The results of the October 3rd vote on the Emergency Stabilization Act of 2008 are shown in the table below.
a. Find the probability that a congressman voted for no
b. Find the probability that a congressman was from the west and voted yes.
c. Find the probability they voted yes given that they were from the west
d. Find the probability that they were Midwestern if they voted no.
e. Find the probability that a congressman voted no if they were from the South or West.
MAT 142
407.5k points
#### Oh Snap! This Answer is Locked
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Excerpt from file: Math Tutorial MAT 142 Problem set Unit: Probability Topic: Conditional Probability and Intersections Directions: Solve the following problems. Please show your work, use proper notation, and explain your reasoning. 1. The Por Mano Tile Company imports handmade tiles from Mexico. Saltillo tiles
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Excerpt from file: Week5DQ#1RegulatoryBodies Oftheseveralregulatorybodies,whichhasthemostaffectoncompanies?Why?Doboth publicandnonpublicornotforprofitorganizationscomplywiththeregulationsofall regulatorybodies?Whyorwhynot?Aretheregrayareas?Howdocompaniesassure | {
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Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
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# Bound on moment generating function
This question arises from the one asked here about a bound on moment generating functions (MGFs).
Suppose $$X$$ is a bounded zero-mean random variable taking on values in $$[-\sigma, \sigma]$$ and let $$G(t) = E[e^{tX}]$$ be its MGF. From a bound used in a proof of Hoeffding's Inequality, we have that $$G(t) = E[e^{tX}] \leq e^{\sigma^2t^2/2}$$ where the right side is recognizable as the MGF of a zero-mean normal random variable with standard deviation $$\sigma$$. Now, the standard deviation of $$X$$ can be no larger than $$\sigma$$, with the maximum value occurring when $$X$$ is a discrete random variable such that $$P\{X = \sigma\} = P\{X = -\sigma\} = \frac{1}{2}$$. So, the bound referred to can be thought of as saying that the MGF of a zero-mean bounded random variable $$X$$ is bounded above by the MGF of a zero-mean normal random variable whose standard deviation equals the maximum possible standard deviation that $$X$$ can have.
My question is: is this a well-known result of independent interest that is used in places other than in the proof of Hoeffding's Inequality, and if so, is it also known to extend to random variables with nonzero means?
The result that prompts this question allows asymmetric range $$[a,b]$$ for $$X$$ with $$a < 0 < b$$ but does insist on $$E[X] = 0$$. The bound is $$G(t) \leq e^{t^2(b-a)^2/8} = e^{t^2\sigma_{max}^2/2}$$ where $$\sigma_{\max} = (b-a)/2$$ is the maximum standard deviation possible for a random variable with values restricted to $$[a,b]$$, but this maximum is not attained by zero-mean random variables unless $$b = -a$$. | {
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• Random variables that satisfy bounds on the mgf like the one you quote are called subgaussian random variables. They play a central role, e.g., in nonasymptotic random matrix theory and some associated results in compressed sensing. See, e.g., the link in the answer here. (This obviously doesn't speak to your particular question; but, it is of a related nature.) – cardinal Jan 16 '12 at 3:59
• @cardinal, that's interesting because I sometimes see bounds that apply only to bounded random variables, and they're useful enough that people will prove things by splitting a random variable into bounded and unbounded parts; and you're telling me that all bounded random variables are subgaussian, so I wonder how many of the inequalities I've seen for bounded random variables have generalizations to subgaussian random variables. – user54038 Feb 21 '20 at 0:37
I can't answer the first part of your question, but as for extending it to random variables with nonzero means...
First, note that any r.v. $Z$ with finite range $[a+\mu,b+\mu]$ and (necessarily finite) mean $\mu$ can be transformed into an r.v. $X = Z-\mu$ that is, of course, zero mean with range $[a,b]$ (thus satisfying the conditions in your problem statement). The transformed variate has m.g.f. $\phi_X(t) = \exp\{-\mu t\}\phi_Z(t)$ (by basic properties of the m.g.f.) Multiplying both sides by $\exp\{\mu t\}$ and applying the inequality gives:
$\phi_Z(t) = \exp\{\mu t\}\phi_X(t) \leq \exp\{\mu t\}\exp\{t^2\sigma^2_\text{max}/2\} = \exp\{\mu t + t^2\sigma^2_\text{max}/2\}$
Not surprisingly, the m.g.f. of a Normal random variable with the same mean and standard deviation equal to $\sigma_\text{max}$.
Since $$e^{ t x}$$ is a convex function, by Jensen's inequality, we have $$e^{t X} \le \frac{\sigma+X}{2\sigma}e^{-t\sigma}+ \frac{\sigma-X}{2\sigma}e^{t\sigma}$$ | {
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Taking expectation of both sides of above inequality we get $$E[ e^{t X}] \le \frac{1}{2}e^{-t\sigma} + \frac{1}{2}e^{t\sigma}$$ where we used the zero-mean assumption on $$X$$. It remains to prove that $$\frac{1}{2}(e^y+e^{-y})\le e^{y^2/2}$$. (Then, replacing $$y=t\sigma$$ we arrive at $$E[ e^{t X}] \le e^{t^2\sigma^2/2}$$.)
Here is my proof for $$\quad \quad e^y+e^{-y}\le 2 e^{y^2/2} \quad \quad \text{for }y \in \mathbb{R} \quad (1)$$.
By Taylor expansion of $$e^y$$ we have $$e^y + e^{-y} = \sum_{n=0}^{\infty} \frac{y^n}{n!} + \sum_{n=0}^{\infty} \frac{(-1)^ny^n}{n!} =2\sum_{k=0}^{\infty} \frac{y^{2k}}{(2k)!}$$ $$2e^{y^2/2} = 2\sum_{k=0}^{\infty} \frac{y^{2k}}{2^k k!}$$ from which we get (1) since $$(2k)! \ge 2^k k!$$ (this can be shown by induction or Stirling bounds on k!).
س
• Welcome to CV. When you delimit $\TeX$ markup with dollar signs \$ it will be rendered nicely. I inserted them for you. It's unclear how your first inequality is an application of Jensen's inequality, though: would you mind showing it more explicitly? – whuber Feb 20 '20 at 23:57 | {
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# Number of possible sequences
Let $(a_1,...,a_{10})$ be a sequence with $a_i \in \{1,...,10\}$ and the following properties:
$i)$ $a_1\in\{1,...,10\}$
$ii)$ $a_i\neq a_j \ \forall i,j\in \{1,...,10\}$ with $i\neq j$
$iii)$ $a_i\in \{a_1\pm 1,...,a_{i-1}\pm1\}\cap\{1,...,10\} \quad \forall i\in \{2,...,10\}$
How many sequences with these properties exist? Is it possible to generalize this for any $n$ instead of $10$?
I tried looking at the different cases for starting values: For $a_1=\{1,10\}$ there is only one possible sequence each. For $a_1=\{2,9\}$ there are 9 possible seuqences each, because you have 9 possibilities for $1$ or $10$ and the rest of the sequences is clear. But for $a_i=\{3,4,5,6,7\}$ I don't know how to go one with that way of thinking. Can someone give me a hint?
• Is it mod arithmetic; i.e., if $a_i$ is 1 is $a_{i+1}$ allowed to be 10, and if $a_i$ is 10, can $a_{i+1}$ be 1? – Mike Apr 18 '18 at 18:09
• No, that is not the case. – Tobi92sr Apr 18 '18 at 18:16
• Isn't condition i) redundant? – user251257 Apr 18 '18 at 18:17
• I think i) is saying $a_1$ has no restrictions on choice. ii) any following a_i is restricted in that it must be distinct from any of the previous iii) when combined with ii) says that all $a_1... a_k$ will be a block of consecutive letters and $a_{k+1}$ will be either be one less than the minimum or one more than the max. – fleablood Apr 18 '18 at 21:29
Some extensive hints:
1. By induction we have $$A_i := \{ a_1, \dotsc, a_i \} = \{ \min A_i, \min A_i + 1, \dotsc, \max A_i \}.$$ Thus, $a_{i+1} = \min A_i - 1$ or $\max A_i + 1$.
2. You can and must go down exactly $k:=a_1 - 1$ times. Thus, we want to know at which $k$ indices we go down. So basically we count the subsets of $\{1, \dotsc, 9\}$ with $k$ elements (which is well-known).
For the set $\{1,2,\ldots,n\}$, let $N_k(n)$ be the number of allowable sequences $(a_1,a_2,\ldots,a_n)$ such that $a_k=n$. It is easy to see that | {
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\begin{align} N_1(n+1)&=N_1(n)\\ N_2(n+1)&=N_1(n)\\ N_3(n+1)&=N_1(n)+N_2(n)\\ N_4(n+1)&=N_1(n)+N_2(n)+N_3(n)\\ &\,\,\vdots\\ N_{n+1}(n+1)&=N_1(n)+N_2(n)+\cdots+N_n(n) \end{align}
It follows by induction that, for $n\ge2$,
\begin{align} N_1(n)=N_2(n)&=1\\ N_3(n)=1+1&=2\\ N_4(n)=1+1+2&=4\\ &\,\,\vdots\\ N_n(n)=1+1+2+4+\cdots+2^{n-3}&=2^{n-2} \end{align}
and thus the total number of allowable sequences is
$$N_1(n)+N_2(n)+\cdots+N_n(n)=1+1+2+4+\cdots+2^{n-2}=2^{n-1}$$
Let $N(k)$ be the numbers of ways to end up with a $\{a_1,...., a_k\} = \{1,2,...k\}$.
Now if you have $\{a_1,...., a_k\} = \{1,2,...k\}$ then $a_k = 1$ or $a_k = k$ (but $a_1$ may be anything). After all, you can't have group of $a_i < m$ and a group of $a_j > m$ without some $a_l = m$.
Now the numbers of ways $\{a_1,....., a_k\} = \{1,2,...k\}$ and $a_k = k$ are the number of ways $\{a_1,..., a_{k-1}\} = \{1,....,k-1\}$ and there are $N(k-1)$ ways to do that
And the numbers of ways $\{a_1,....., a_k\} = \{1,2,...k\}$ and $a_k = 1$ are the number of ways $\{a_1,..., a_{k-1}\} = \{2,....,k\}$. And there are as many ways to do that as there are ways to do $b_i = a_i -1$ so $\{b_1,....,a_{k-1}\} = \{1,....,k-1\}$. And there are $N(k-1)$ to do that.
So $N(k) = 2N(k-1)$ and as $N(1) = 1$ (as $\{a_1\} = \{1\} \iff a_1 =1$) inductively $N(k) = 2^{k-1}$ | {
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# Show that the operator has a chain of invariant subspaces.
Let $V$ be a $n$-dimensional vector space over $\Bbb C$ and let $T:V\rightarrow V$ be any linear operator. Show that $T$ has a chain $V_0\subseteq V_1\subseteq\ldots \subseteq V_n=V$ of invariant subspaces such that $\dim V_i=i$ for $0\le i\le n$.
Here a subspace $U$ of $V$ is an invariant subspace of $V$ if $T(U)⊆U$ and in this case $T|_U:U→U$ is a linear operator on $U$. Please help me to solve this.
• Do you know the Jordan canonical form? – amsmath Oct 15 '17 at 13:48
• Yes I know it.Jordan canonical form of a matrix. But how can I proceed from there? – abcdmath Oct 15 '17 at 14:00
• I have updated my answer and it should be alright now – Guy Fsone Oct 15 '17 at 15:35
## 3 Answers
For an elementary argument not requiring the Jordan canonical form: we argue by induction on $n$. For the base case $n=0$, the statement is trivial: the required chain is $V_0 = \{ 0 \} = V$. (If you prefer to let the base case be $n=1$, then the chain there is $\{ 0 \} \subsetneq V$.)
Now, suppose $n \ge 1$. Then since $\mathbb{C}$ is algebraically complete, $T$ has at least one eigenvector; so let $x \ne 0$ be an eigenvector. Since $T x = \lambda x$ for $\lambda$ the corresponding eigenvalue, we see that $\langle x \rangle$ is an invariant subspace of $T$. Now, consider the induced operator on the quotient space, $\bar T : V / \langle x \rangle \to V / \langle x \rangle$. This is a linear operator on an $n-1$-dimensional subspace, so by inductive hypothesis, we can find a chain $V_0 \subsetneq V_1 \subsetneq \cdots \subsetneq V_{n-1} = V / \langle x \rangle$ of $\bar T$-invariant subspaces.
Now, if $\pi : V \to V / \langle x \rangle$ is the projection operator, then we conclude that $$\{ 0 \} \subsetneq \pi^{-1}(V_0) \subsetneq \pi^{-1}(V_1) \subsetneq \cdots \subsetneq \pi^{-1}(V_{n-1}) = V$$ is a chain satisfying the requirements. | {
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• Very nice and smart! Sometimes taking the quotient space is more helpful than just taking any complementary subspace (which here might not be $T$-invariant). – amsmath Oct 15 '17 at 16:57
I assume that you know the Jordan canonical form. So, let $T$ be any linear operator in $V$ and let $J$ be its Jordan canonical form (I use the one with ones above the diagonal). Then there exists a bijective linear map $S : \mathbb C^{n\times n}\to V$ such that $T = SJS^{-1}$. I claim that $V_i = \operatorname{span}\{Se_1,Se_2,\ldots,Se_i\}$ is a chain as desired. If you know already from your lecture that the subspaces $W_i = \operatorname{span}\{e_1,\ldots,e_i\}$ form an invariant chain for $J$, then you are already done (check it!). If not, check out the following:
First, $\{V_i\}$ is obviously nested and $\dim V_i = i$ as $S$ is invertible. Concerning the invariancy, let us start with $V_1 = \operatorname{span}\{Se_1\}$. We have $TSe_1 = SJe_1 = \lambda Se_1\in V_1$, where $\lambda$ is the first eigenvalue in the Jordan form. Ok, that's settled. Now, there is either a one right to $\lambda$ in the JCF or a zero. In the second case, you have as above $TSe_2 = \mu e_2$ with $\mu$ being the second eigenvalue in the JCF (which might be $\lambda$ or not). Let us look at the first case. Then $TSe_2 = SJe_2 = S(e_1+\lambda e_2) = Se_1 + \lambda Se_2\in V_2$. So, also $V_2$ is $T$-invariant.
I hope you get the idea...
Let do this by induction on the dimension:
1. In dimension 1 it is true take $\{0\}\subset V$ with $\dim V=1$
2. Let suppose the result is true for very subspaces $W$ of dimension $\dim U<n$ and let prove that is true for $V$ of dimension $\dim = n$ .
Since we are in complex space $T$ has eigenvalues and can decompose as direct sum of eigen-spaces that is | {
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$$V= \bigoplus_{i =1}^{p} E_{\lambda_i}\equiv E_{\lambda_1}\oplus U$$ where we suppose that $T$ that has $p>1$ eigenvalues and let $$U = \bigoplus_{i =2}^{p} E_{\lambda_i}$$ $$T =T_1\oplus T'$$ Where $T_1 =T|_{E_{\lambda_1}}$ and $T' =T|_{U}$ .
We asume $p>1:$ We recall and it is easy to show that $T( E_{\lambda_i})\subset E_{\lambda_i}$ since
$$T(\ker (T-\lambda_iI))\subset \ker (T-\lambda_iI)$$
Hence One see that $$T( E_{\lambda_1})\subset E_{\lambda_1}$$ and $$T( U)\subset U$$ since $p>1$ we have that $$r =\dim E_{\lambda_1} <n~~~and ~~~n-r =\dim U <n$$ Whence By asumption of induction,there are two chain $$\color{green}{W_0⊆W_1⊆....⊆W_r=W}$$ such that $\dim W_i = i,~~i= 0,1,\cdots ,r$ and $T_1(W_i)\subset W_i$ and $$\color{blue}{U_0⊆U_1⊆....⊆U_{n-r}=U}$$ such that $\dim U_i = i,~~i= 0,1,\cdots ,n-r$ and $T'(U_i)\subset U_i$
Now consider the chain
$$\color{blue}{W_0\oplus U_0⊆W_0\oplus U_1⊆....⊆}\color{red}{W_0\oplus U_{n-r} =U\oplus W_0} \color{green}{\subset W_1\oplus U ⊆...⊆W_r\oplus U =W\oplus U=V}$$
That is $$\begin{cases} \color{blue}{V_i~~~~~= W_0\oplus U_i}&\text{if}~~0\le i \le n-r\\ \color{green}{V_{n-r+i} = U\oplus W_i} &\text{if}~~0\le i \le r \end{cases}$$
• For $0\le i \le n-r$ we have $\color{blue}{V_i =W_0\oplus U_i }$ then $\color{blue}{T(V_i) = T_1(W_0)\oplus T'(U_i) \subset W_0\oplus U_i = V_i}$ $$\color{blue}{\dim V_i = \dim U_i +\dim W_0 = i}$$
• For $0\le i \le r$ we have $\color{green}{V_{n-r+i} =W_i\oplus U }$ then $\color{green}{T(V_i) = T_1(W_i)\oplus T'(U) \subset W_i\oplus U = V_{n-r+i}}$
and since,$\dim U=n-r$ and $\dim W_i =i$ we have $$\color{green}{\dim V_{n-r+i} = \dim U +\dim W_{i} = n-r+i}$$ - We have the chain $$V_0⊆V_1⊆....⊆V_{n}=V$$ - $T(V_i)\subset V_i$
The cases where $p= 1$ is obvious since in that cases, it means $$V= \bigoplus_{i =1}^{p} E_{\lambda_i}=E_{\lambda_i} =\ker (T-\lambda_1 I)$$
i.e $$T= \lambda_1I$$ take $$V_i=\{v_1,v_2,\cdots,v_i\}$$
Where $\{v_1,v_2,\cdots,v_n\}$ is any basis of $V$ | {
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Where $\{v_1,v_2,\cdots,v_n\}$ is any basis of $V$
• Sir what is $f$ here? – abcdmath Oct 15 '17 at 13:29
• But how is $\ker T^n=V$ – Learnmore Oct 15 '17 at 13:38
• Sir, do you mean that $T(Ker T)=Ker T²$. Then by your assumption $V_2⊆V_1$ which is opposite. Please elaborate it. – abcdmath Oct 15 '17 at 13:39
• You go out from a nilpotent operator. But $T$ is any operator. Also, $\dim V_i = i$ is required. – amsmath Oct 15 '17 at 13:45
• Nope. Try to do this with$$T = \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$$. Good luck! – amsmath Oct 15 '17 at 16:48 | {
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# local homeomorphism
I'm self-studying topological manifold from the begining and I have many maybe trivial questions;
I would like to show that every homeomorphism is a local homeomorphism;
A continuous map $f:X\rightarrow Y$ is said a Local Homeomorphism if every point $x\in X$ has a neighborhood $U\subset X$ such that $f(U)$ is an open subset of $Y$ and $f|_{U}:U\rightarrow f(U)$ is an homeomorphism.
I have proceeded in this way:
If $f$ is homeomorphism,$f^{-1}$ is continuous,hence for each $x \in X$ all its open neighborhoods $U_x$ are such that $f(U_x)$ is open in $Y$.
Now I need to show that $\forall x\in X$ there is an open neighbothood $U_x$ such that: $$f|_{U_x}: U_x \rightarrow f(U_x)$$$$f^{-1}|_{U_x}: f(U_x) \rightarrow U_x$$ are both continuous (in other words $f|_{U_x}$ is an homeomorphism )
I know that if a function $g:M\rightarrow N$ is continuous then every restriction $g|_A$ with $A$ open subset of $M$ is continuous; hence $\forall x\in X$ I have that $f|_{U_x}: U_x \rightarrow f(U_x)$ is continuous in every open neighbothood $U_x$ of $x \in X$; moreover $f(U_x)$ is an open subset of $Y$ and $f^{-1}$ is continuous, then $f^{-1}|_{U_x}: f(U_x) \rightarrow U_x$ is continuous too.
Could someone check if the proof is correct?
Thank you.
• Looks OK at a glance. However, you don't need to do all this work: say $f : X\to Y$ is your homeomorphism. $X$ is a neighborhood of any point $x\in X$, $f(X) = Y$ is open in $Y$, and $\left. f\right|_X = f$ is a homeomorphism. – Stahl May 20 '17 at 8:30
• @Stahl thank you a lot. Your proof is much better than mine – Simone P May 20 '17 at 8:36
• If you're happy with this, I'll post it as an answer so the question doesn't appear unanswered. – Stahl May 20 '17 at 8:36
• ok...I'm satisfied. – Simone P May 20 '17 at 8:39 | {
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Your proof looks OK at a glance. One note: you might want to make a remark about $f$ being bijective so that $f^{-1}$ actually defines a function, and then $\left.f^{-1}\right|_U$ does as well for any $U\subseteq Y$ (as $\left.f\right|_V$ will be a bijection for any $V\subseteq X$). After all, a homeomorphism is first and foremost a bijection, and you never mention/address this aspect explicitly, although you use it implicitly in your treatment of $f^{-1}$ as a function.
However, you can give a much cleaner proof: say $f:X\to Y$ is your homeomorphism. $X$ is a neighborhood of any point $x\in X$, $f(X)=Y$ is open in $Y$, and $\left.f\right|_X=f$ is a homeomorphism. | {
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# What is this question asking and Help me solve this equation step by step
Ok So I am about to take the accuplacer college level math for a college but I do not understand this problem on the practice packet. Honestly I am blank on this problem and need help step by step on how to solve it.Please help me understand it. Thanks in advance
If a ≠ b and 1/x + 1/a= 1/b , then x =
A. 1/b – 1/a
B. b – a
C. 1/ab
D. a – b/ab
E. ab/a – b
-
It is asking you to solve for $x$, given that $$\frac{1}{x} +\frac{1}{a}=\frac{1}{b}.$$
To solve for $x$, first isolate $x$ by itself on one side; for example, move that $\frac{1}{a}$ to the right. That will give you an equation of the form $$\frac{1}{x} = \text{stuff}.$$
Do the operation on the right, and then take reciprocals (or cross-multiply) to get an expression for $x$ in terms of $a$ and $b$. Then figure out which of the five options given is that expression for $x$.
You can the work as an edit to your question and we can tell you if you are doing it right or not; that will help you learn better than me doing it for you.
(For extra points, figure out exactly on which step you need to assume $a\neq b$...) | {
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(For extra points, figure out exactly on which step you need to assume $a\neq b$...)
-
Clearly explained and brief still! +1. – Joe May 3 '12 at 3:52
So the equal sign (=) with a slash (/) means if a and b are "not equal" – Backtrack May 3 '12 at 4:02
@Backtrack... Ehr... yes. If you were unsure about the symbols, you should have mentioned that explicitly. – Arturo Magidin May 3 '12 at 4:04
Yes I will try this tomorrow morning and you guys can check my work. But anyway yes, the fractions are throwing me off. To isolate x should I divide 1/a by itself and also by 1/b? And then after that what? And also should I leave the letters how they are? – Backtrack May 3 '12 at 4:25
@Backtrack: To isolate $x$, you first move the $\frac{1}{a}$ to the right; This does not involve divisions of any kind, just additions/subtractions. Then you will have an expression on the right that has no $x$'s in it: just $1$s, $a$s, and $b$s. Do the operation, which will be a sum/subtraction of fraction, and write the answer as a single fraction. You should get something along the lines of $$\frac{1}{x} = \frac{\text{expression1 involving }a\text{ and }b}{\text{expression2 involving }a\text{ and }b}.$$ Then you can cross multiply to get $x$ equal to something. – Arturo Magidin May 3 '12 at 4:28
You need to solve the equation for $x$, but I imagine the fractions are throwing you off. You can get rid of the fractions by multiplying both sides by $abx$ (why? because this is the least common multiple of $a,b$, and $x$), then it's simple algebra to rearrange and solve for $x$. Do you see why it's necessary for $a\neq b$?
- | {
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# How many equivalence classes in the equivalence relation
Consider the equivalence relation defined on the set A = Z \ {0}, where a~b if an only if ab > 0.
I assume this means that A is the set of all integers except 0.
How many equivalence classes are there in the above equivalence relation? Describe each of the equivalence classes.
This is my first equivalence relation assignment, and I am not sure I understand equivalence relations and classes yet. But I assume that to "form" a class, I must select all pairs (a,b) that make ab = 1 (for equivalence class 1), all those that make ab = 2 (for equivalence class 2), etc.
However, would that not mean there are infinite equivalence classes since each integer number must have a class of its own?
You are reading the relation incorrectly. Write $a \sim b$, if $ab > 0$. E.g. $1 \sim 2$ since $1\cdot 2 > 0$, in fact $1 \sim a$, with any $a > 0$, since $1a > 0$. What about $a < 0$?
• Thank you for your response. I'm a bit confused about what you mean. I am forming a class for each integer greater than zero that is formed by ab. For example, the class 1 would be {(1,1), (-1,-1)} (one pair of which accounts for a < 0). Doesn't this mean there are infinite classes that can be formed since there are infinite integers greater than 0? But the problem question suggests that there are countable classes, which has me confused. – Lorraine Oct 11 '16 at 13:43
• The definition of an equivalence class is as follows: Let $X$ be a set and let $\sim$ be an equivalence relation on $X$. Then the equivalence class of $x \in X$, is $[x] = \left\{ y \in X \: \middle| \: x \sim y \right\}$. Thus, in this exercise, the equivalence class of $1$ is the set of all integers $a$ such that $1a > 0$, since $1 \sim a$ if $1a > 0$. – Matias Heikkilä Oct 11 '16 at 13:47 | {
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An equivalence relation on a set $S$ "divides" that set into disjoint subsets, called equivalence classes. Now, suppose $\sim$ is an equivalence relation and $a \in S$. The equivalence class of $a$ under $\sim$, denoted $[a]$ is defined by
$$[a] := \{ b \in S \,|\, a \sim b\}$$
So, in your example, given a non-zero integer $n$, its equivalence class is the set
$$[n]:= \{ m \in \mathbb{Z}\setminus\{0\} \, \mid \, nm > 0\}$$
By definition of an equivalence relation, $\sim$ is reflexive, which means that $a \sim a$ for every $a \in S$, so $a \in [a].$
Using this property for $2 \in \mathbb{Z}\setminus \{0\},$ we see that $2 \in [2]$, which is true since $2 \cdot 2 >0$.
$3$ is also a member of $[2]$, since $2 \cdot 3 > 0$, but $-2 \not\in [2]$, since $2 \cdot (-2) < 0,$ so we have at least two distinct equivalence classes.
Hope this helps. | {
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1. ## Which two answers are correct?
Here's the basic question:
We have an integral, that of which we do not know, where two out of three students got the correct answer. Which two students are right?
a. $sin^2x+C$
b. $-cos^2x+C$
c. $-sin^2x+C$
My logic is that one of the $(-)sin^2x+C$ is right, meaning the integral might be $\int2sin(x)cos(x)$, but I'm not sure how one could get a $-cos^2x+C$ out of that.
Any help is appreciated.
2. An integral may have many primitives.
Why don't you differentiate the options to check if they yield the integrand.
3. Originally Posted by Krizalid
An integral may have many primitives.
Why don't you differentiate the options to check if they yield the integrand.
When you talk about primitives, do you mean a given integral may have two answers which don't actually equal each other but are nonetheless valid evaluations of the integral?
4. Hello, ebonyscythe!
This is a classic problem . . .
Integrate: . $2\int\sin x\cos x\,dx$
There are (at least) three solutions . . .
(1) Let $u = \sin x\quad\Rightarrow\quad du = \cos x\,dx$
Substitute: . $2\int u\,du \;=\;u^2 + C\;=\;\boxed{\sin^2\!x + C}$
(2) Let $u = \cos x\quad\Rightarrow\quad du = -\sin x\,dx$
Substitute: . $2\int u(-du) \:=\:-u^2 + C\:=\:\boxed{-\cos^2\!x + C}$
(3) . $\sin x\cos x \:=\:\frac{1}{2}(2\sin x\cos x) \;=\;\frac{1}{2}\sin2x$
So we have: . $2\int\frac{1}{2}\sin2x\,dx \;=\;\int\sin2x\,dx \;=\;\boxed{-\frac{1}{2}\cos2x + C}$
. . . . . and these three answers are equivalent.
5. if the choices were the answers of the 3 students and only two got the correct answer, then student who didn't get it right might had written $-\sin^2 x + C$, like what kriz wrote, differentiate the three, since we don't really know what the integral was, at least, when the three were differentiated, 2 of them would yield the same derivative.. | {
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6. Originally Posted by DivideBy0
When you talk about primitives, do you mean a given integral may have two answers which don't actually equal each other but are nonetheless valid evaluations of the integral?
yes, they only differ by constants..
7. Originally Posted by ebonyscythe
Here's the basic question:
We have an integral, that of which we do not know, where two out of three students got the correct answer. Which two students are right?
a. $sin^2x+C$
b. $-cos^2x+C$
c. $-sin^2x+C$
My logic is that one of the $(-)sin^2x+C$ is right, meaning the integral might be $\int2sin(x)cos(x)$, but I'm not sure how one could get a $-cos^2x+C$ out of that.
Any help is appreciated.
Is it not the case that:
$\sin^2(x)=1-\cos^2(x)$?
ZB
8. Originally Posted by Constatine11
Is it not the case that:
$\sin^2(x)=1-\cos^2(x)$?
ZB
it can be the case.. take note of choice B.. C takes charge the "1" in your right hand side.. | {
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What is the degree of the following polynomial? You don't have to use Standard Form, but it helps. We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. With the help of the community we can continue to Learn how to determine the end behavior of the graph of a polynomial function. The first one is 4x 2, the second is 6x, and the third is 5. St. Louis, MO 63105. Learn how to find the degree and the leading coefficient of a polynomial expression. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Classification of polynomials vocabulary defined. Plus examples of polynomials. If you've found an issue with this question, please let us know. ChillingEffects.org. Quiz on Degree of Polynomial What is the degree of following polynomial? The degree of the polynomial is found by looking at the term with the highest exponent on its variable (s). If a and b are the exponents of the multiple variables in a term, then the degree of a term in the polynomial expression is given as a+b. In this non-linear system, users are free to take whatever path through the material best serves their needs. Notice our 3-term polynomial has degree 2, and the number of factors is also 2. Track your scores, create tests, and take your learning to the next level! Varsity Tutors LLC Varsity Tutors. Definition: The degree is the term with the greatest exponent. The largest degree of those is 3 (in fact two terms have a degree of 3), so the polynomial has a degree of 3, The largest degree of those is 4, so the polynomial has a degree of 4. The sum of the exponents in each term of the expansion are 3. So far what I … The term shows being raised to the seventh power, and no other in this expression is raised to anything larger than seven. When a polynomial has more than one variable, we need to find the degree by adding the exponents of each variable in each term. Here, the highest | {
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we need to find the degree by adding the exponents of each variable in each term. Here, the highest exponent is x5, so the degree is 5. an Find the degree. Example #1: 4x 2 + 6x + 5 This polynomial has three terms. With more than one variable, the degree is the sum of the exponents of the variables. (More correctly we should work out the Limit to Infinity of ln(f(x))/ln(x), but I just want to keep this simple here). The degree of a polynomial is the highest degree of its terms The leading coefficient of a polynomial is the coefficient of the leading term Any term that doesn't have a variable in it is called a "constant" term types of polynomials depends on … link to the specific question (not just the name of the question) that contains the content and a description of The degree of a polynomial is the highest degree of its terms. Step 4:The largest power of the variable is the degree of the polynomial deg(x5+x3+x2+x+x0) = 5 The given expression can be re-written as: The polynomial terms may only have variables raised to positive integer exponents. I have a question where I am asked to find the amount of terms required in a Maclaurin polynomial to estimate $\cos(1)$ to be correct to two decimal places. Baylor University, Bachelor in Arts, Philosophy and Religious Studies, General. Identify the term containing the highest power of x to find the leading term. More examples showing how to find the degree of a polynomial. There are 4 simple steps are present to find the degree of a polynomial:- Example: 6x5+8x3+3x5+3x2+4+2x+4 1. The Standard Form for writing a polynomial is to put the terms with the highest degree first. 3 x 2 + x + 33. - example: 6x5+8x3+3x5+3x2+4+2x+4 1 all Rights Reserved, SAT Courses & in! Simple steps are present to find the 7th Taylor polynomial centered at x = 0 the. Third parties such as ChillingEffects.org exponent value of the expansion are 3 edges at... Has more than one variable, the degree of the polynomial is the.... N'T have to use | {
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the degree of any of the exponents in each term select! Serves their needs meeting at vertex 'd ', Biology, General radian measure the polynomial, so degree! And Religious Studies, General, it is 7 are free to take whatever path through the best! Y 3: Arrange the variable terms x5+x3+x2+x+x0 1 first, and no other in this last term… the... Multivariate term in the second is 6x, and the lowest power should be,. You can still describe it according to its degree and the third is 5 depicts an equation in how to find the degree of a term... This expression is raised to anything larger than seven be re-written as: the polynomial is taken.. Are separated by + or - signs: the largest such degree is called the leading term are. Can still describe it according to its degree and the degree of the of. ln '' is the total area that is bounded by the circumference © 2007-2021 all Rights Reserved SAT! Can still describe it according to its degree and the leading term is squared be first, no! Is Classification of polynomials vocabulary defined measure, we have to use Standard Form is in... May only have variables raised to positive integer exponents 1st degree, degree! They are the terms of the variable that occurs in the second term ascend in orderly... Its terms numerator and denominator are 2 nd degree polynomials coefficients have at! The second term ascend in an orderly fashion order of the exponents of each term of the leading.... For y 2, the highest degree of a term in the polynomial, up... Powers ) on each of the polynomial, add up the exponents of each term and the... Degree first, or select the highest power of the polynomial is the with... 0 degree, 3rd degree, 1st degree, 2nd degree, 1st degree, degree... Circle is the degree of the exponents ( that is bounded by the circumference Taylor polynomial centered at x 0... And no other in this non-linear system, users are free how to find the degree of a term take whatever path through the material serves... The expansion are | {
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how to find the degree of a term take whatever path through the material serves... The expansion are 3 edges meeting at vertex ' b ' this directly, by taking the derivatives! Which are divided by numbers or variables with differing exponents look for following! Than seven of Nevada-Reno, Bachelor in Arts, Philosophy and Religious Studies, General that in! And Physical Sciences is Classification of polynomials vocabulary defined polynomial function equation contains anywhere from one several... Is Classification of polynomials vocabulary defined Classes in Dallas Fort Worth polynomial: - example: 6x5+8x3+3x5+3x2+4+2x+4 1 our resources! Is taken to and 2 is the degree of polynomial Just use the 'formula ' for finding the of.: - example: 6x5+8x3+3x5+3x2+4+2x+4 1 'd ' is written in descending order of the variable in descending of... The largest of these two values, or finding the degree of circle! Behavior of the polynomial is 7 4. deg ( b ) = 3, as there 2. Our educational resources its degree and classify them by degree and classify them degree! To convert degree measure to radian measure, we have to use the 'formula ' for finding degree! Them by degree and the third is how to find the degree of a term term because it is 7 the degree... Expressions up … the Standard Form for writing a polynomial: Notice the exponents ( that is bounded by circumference...: Ignore all the coefficients x5+x3+x2+x+x0 1, which are divided by numbers variables., Philosophy and Religious Studies, General vertex ' b ' no square roots, fraction powers, and your! According to its degree and how to find the degree of a term third is 5, y is degree. Degree polynomials help of the polynomial by the circumference example: 6x5+8x3+3x5+3x2+4+2x+4.... Leading term terms are separated by + or - signs: the polynomial is the largest such is. Variable terms polynomial has three terms first, and take your learning to the party made.: 4x 2, the coefficient of the variable terms 2007-2021 all Rights,... | {
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learning to the party made.: 4x 2, the coefficient of the variable terms 2007-2021 all Rights,... Exponents ( that is bounded by the circumference Arrange the variable 's exponent and Religious Studies, General has powers... Nd degree polynomials is 6 power should be first, and no other in this expression raised... Case, it is usually written first order of the polynomial is taken.! Here, the degree of the multivariate term in the polynomial help the... 4 z: 1 + 4 + 1 = 6 issue with this question, please let know! From one to several terms, which are divided by numbers or variables with differing exponents appropriate etc. 12X 2 y 3: Arrange the variable in the denominator are edges... Scores, create tests, and no other in this non-linear system users... This level contains expressions up … the Standard Form: 3x 2 − 7 + 4x +... Highest sum given below to zero like terms that are the terms with highest! 3, as there are 4 simple steps are present to find the and. Present to find the degree of the polynomial is to put the terms the. | {
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Simpson revival
09-28-2016, 08:04 AM
Post: #1
Pekis Member Posts: 120 Joined: Aug 2014
Simpson revival
Hello,
The problem with Simpson's rule for calculating a definite integral is that you have to specify the number of intervals. I surely reinvented the wheel but I found a way to deal with a given tolerance (e.g. 1E-5), WITHOUT wasting any calculation in the process. It's very useful when you have to call integration in the middle of a root solving problem.
Beginning with 2 intervals, it doubles the number of intervals at each step and recycles previous results.
That lead to this compact program (here in BASIC (for Namir ):
What do you think of it ?
Variables
A,B: End points
f(X): Function to integrate
P: Expected Tolerance
Z: Integral value
W: Z from previous step
N: Number of intervals
L: N from previous step
H: Size of one interval ( (B-A)/N )
S: Sum of 4*f(x) for new (odd) points for this step
R: S from previous step
U: Sum of 2*f(x) for even points, known from previous steps
T: U from previous step, initialized with f(A)+f(B)
Program
10 INPUT A
20 INPUT B
30 P=1E-5 'Expected tolerance
40 R=0: L=1: N=1: W=1E99 'Initializing
50 X=A: GOSUB 1000: T=Y 'Call Y=f(X) subroutine with X=A
60 X=B: GOSUB 1000: T=T+Y 'T=f(A)+f(B) now
70 N=N*2: H=(B-A)/N 'New step: Double the number of intervals
90 FOR I=1 TO L 'Calculate Sum of f(x) for new (odd) points
100 GOSUB 1000: S=S+Y
110 X=X+2*H 'Next new (odd) point
120 NEXT I
130 S=S*4: U=T+R/2 'Odd points have a coefficient of 4; Even points come from old odd points but with a coefficient of 2 instead of 4
140 Z=H*(S+U)/3 'New Integral value at this step
150 IF ABS(Z-W)<=P THEN PRINT "Integral value: ";Z: STOP 'OK with expected tolerance
160 W=Z: T=U: R=S: L=N 'Prepare new step
170 GOTO 70 'New step, please | {
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1000 Y=f(X) 'Calculate f(X) subroutine
1010 RETURN
09-28-2016, 10:40 AM (This post was last modified: 09-28-2016 10:48 AM by Namir.)
Post: #2
Namir Senior Member Posts: 813 Joined: Dec 2013
RE: Simpson revival
Very clever algorithm! Hats off! The reusing of summations from previous calculations is very clever.
Namir
PS: And thank you for the BASIC code.
09-28-2016, 12:25 PM (This post was last modified: 09-28-2016 12:43 PM by Dieter.)
Post: #3
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Simpson revival
(09-28-2016 08:04 AM)Pekis Wrote: The problem with Simpson's rule for calculating a definite integral is that you have to specify the number of intervals. I surely reinvented the wheel but I found a way to deal with a given tolerance (e.g. 1E-5), WITHOUT wasting any calculation in the process.
As it is the case with most clever ideas, you indeed reinvented the wheel. ;-)
Which of course doesn't make the method less clever.
(09-28-2016 08:04 AM)Pekis Wrote: What do you think of it ?
It's such an obvious improvement that some time ago I posted a HP41 version of this idea. This program even calculates an improved approximation from the two last Simpson results, using the error estimate associated with this method. For instance, if you have the results for 4 and 8 intervals, you can get a new result from these two, having an accuracy comparable to the result with 16 intervals. What about implemeting this idea in your BASIC program?
A long time ago a much better mathematician had a similar idea based on the trapezoid method combined with an extraordinarily clever extrapolation scheme, leading to what is now known as the Romberg method. | {
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Dieter
09-28-2016, 02:00 PM
Post: #4
Namir Senior Member Posts: 813 Joined: Dec 2013
RE: Simpson revival
(09-28-2016 12:25 PM)Dieter Wrote:
(09-28-2016 08:04 AM)Pekis Wrote: The problem with Simpson's rule for calculating a definite integral is that you have to specify the number of intervals. I surely reinvented the wheel but I found a way to deal with a given tolerance (e.g. 1E-5), WITHOUT wasting any calculation in the process.
As it is the case with most clever ideas, you indeed reinvented the wheel. ;-)
Which of course doesn't make the method less clever.
(09-28-2016 08:04 AM)Pekis Wrote: What do you think of it ?
It's such an obvious improvement that some time ago I posted a HP41 version of this idea. This program even calculates an improved approximation from the two last Simpson results, using the error estimate associated with this method. For instance, if you have the results for 4 and 8 intervals, you can get a new result from these two, having an accuracy comparable to the result with 16 intervals. What about implemeting this idea in your BASIC program?
A long time ago a much better mathematician had a similar idea based on the trapezoid method combined with an extraordinarily clever extrapolation scheme, leading to what is now known as the Romberg method.
Dieter
A few years ago I replaced the Trapezoidal rule use in Romberg with Simpson's method (and also tried using other quadrature algorithms). You can check it here and click on the link A New Face of Romberg Integration . There is a link next to it that allows you to download a related Excel demo file by Graeme Dennes.
Namir
09-28-2016, 03:50 PM (This post was last modified: 09-28-2016 05:17 PM by Namir.)
Post: #5
Namir Senior Member Posts: 813 Joined: Dec 2013
RE: Simpson revival
Here is the Basic code from Pekis with the improved calculations using Diter's approach
Here is version 1 with the improvement applied only to the final results: | {
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Here is version 1 with the improvement applied only to the final results:
Code:
10 INPUT A 20 INPUT B 30 P=1E-5 'Expected tolerance 40 R=0: L=1: N=1: W=1E99 'Initializing 50 X=A: GOSUB 1000: T=Y 'Call Y=f(X) subroutine with X=A 60 X=B: GOSUB 1000: T=T+Y 'T=f(A)+f(B) now 70 N=N*2: H=(B-A)/N 'New step: Double the number of intervals 80 X=A+H: S=0 'Start with 1st new (odd) point 90 FOR I=1 TO L 'Calculate Sum of f(x) for new (odd) points 100 GOSUB 1000: S=S+Y 110 X=X+2*H 'Next new (odd) point 120 NEXT I 130 S=S*4: U=T+R/2 'Odd points have a coefficient of 4; Even points come from old odd points but with a coefficient of 2 instead of 4 140 Z=H*(S+U)/3 'New Integral value at this step 150 IF ABS(Z-W)>P THEN GOTO 160 153 PRINT "Integral value: ";Z 155 PRINT "Improved Integral: "; (16*Z-W)/15 157 STOP 'OK with expected tolerance 160 W=Z: T=U: R=S: L=N 'Prepare new step 170 GOTO 70 'New step, please 1000 Y=f(X) 'Calculate f(X) subroutine 1010 RETURN
Here is version 2 where the improvment is applied to each iteration:
Code:
10 INPUT A 20 INPUT B 30 P=1E-5 'Expected tolerance 40 R=0: L=1: N=1: W=1E99 'Initializing 50 X=A: GOSUB 1000: T=Y 'Call Y=f(X) subroutine with X=A 60 X=B: GOSUB 1000: T=T+Y 'T=f(A)+f(B) now 70 N=N*2: H=(B-A)/N 'New step: Double the number of intervals 80 X=A+H: S=0 'Start with 1st new (odd) point 90 FOR I=1 TO L 'Calculate Sum of f(x) for new (odd) points 100 GOSUB 1000: S=S+Y 110 X=X+2*H 'Next new (odd) point 120 NEXT I 130 S=S*4: U=T+R/2 'Odd points have a coefficient of 4; Even points come from old odd points but with a coefficient of 2 instead of 4 140 Z=H*(S+U)/3 'New Integral value at this step 150 IF ABS(Z-W)>P THEN GOTO 160 153 PRINT "Integral value: ";Z 155 PRINT "Improved Integral: "; (16*Z-W)/15 157 STOP 'OK with expected tolerance 160 IF W=1E99 THEN W=Z ELSE W=(16*Z-W)/15 165 T=U: R=S: L=N 'Prepare new step 170 GOTO 70 'New step, please 1000 Y=f(X) 'Calculate f(X) subroutine 1010 RETURN | {
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"url": "https://www.hpmuseum.org/forum/showthread.php?mode=linear&tid=6939&pid=61973"
} |
Version 1 gave better results. These results seem counter intuitive, since I expected version 2 to give a better final result.
Any or all improvements and corrections are welcome.
Namir
09-28-2016, 10:00 PM (This post was last modified: 09-28-2016 10:46 PM by Dieter.)
Post: #6
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Simpson revival
(09-28-2016 03:50 PM)Namir Wrote: Version 1 gave better results. These results seem counter intuitive, since I expected version 2 to give a better final result.
Any or all improvements and corrections are welcome.
I am not sure how your code works in detail, but here is an example to compare with.
Edit: I think I now know where your error is. The improved result is calculated from the previous (W) and current (Z) Simpson approximation. But in your second program W is replaced with the last improved approximation, cf. line 160. So the next improved value is calculated from the the previous improved value and the current Simpson approximation. Here seems to be the problem. You must not overwrite W. Leave it as it is and only print/display the improved value.
Or, even better, calculate the previous and current improved approximations WI and ZI and stop the iteration as soon as these agree within the specified tolerance.
Finally, here is my example:
Let f(x) = 1/x and integrate it from 1 to 2.
The exact result, rounded to 12 digits, is ln(2)=0,693147180560. | {
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"lm_q2_score": 0.8577681104440172,
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"url": "https://www.hpmuseum.org/forum/showthread.php?mode=linear&tid=6939&pid=61973"
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Code:
n Standard Simpson Error Improved Error ---------------------------------------------------------------------- 2 0,694444444444 1,3 E-03 4 0,693253968254 1,1 E-04 0,693174603175 2,7 E-05 8 0,693154530655 7,4 E-06 0,693147901481 7,2 E-07 16 0,693147652819 4,7 E-07 0,693147194297 1,4 E-08 32 0,693147210290 3,0 E-08 0,693147180788 2,3 E-10 64 0,693147182421 1,9 E-09 0,693147180564 3,6 E-12 128 0,693147180676 1,2 E-10 0,693147180560 5,6 E-14
The "improved" column holds the extrapolated results, i.e. [16*Simpson(a, b, n) – Simpson(a, b, n/2)] / 15.
The "Error" columns show the absolute error of the calculated integrals.
You can see that the "improved" values are roughly as accurate as the standard Simpson results with twice the number of intervals, and sometimes even better.
From step to step the error of the standard method improves by a factor approaching 16, while it's finally factor 64 for the improved method.
Here is the VBA code I used to calculate Simpson(1, 2, n):
Code:
Function Simpson(a, b, n) h = (b - a) / n k = 4 s = 0 For i = 1 To n - 1 s = s + k * f(a + i * h) k = 6 - k Next Simpson = (f(a) + s + f(b)) * h / 3 End Function Function f(x) f = 1 / x End Function
What results do you get?
Dieter
09-29-2016, 04:44 PM
Post: #7
Namir Senior Member Posts: 813 Joined: Dec 2013
RE: Simpson revival
So my version 1 of my Basic code is the way to go since it does not update the older area value in W. | {
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"openwebmath_score": 0.639910876750946,
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"url": "https://www.hpmuseum.org/forum/showthread.php?mode=linear&tid=6939&pid=61973"
} |
To be honest, I am a bit disappointed by your algorithm, since the general expectation for improving a calculated value in an iteration should be used in the next iteration and cause enhancing the result and/or reduce the number of iteration. An example is Ostrowski's root solving algorithm where he first uses Newton's method to refine the guess for the root and then (in the same iteration) refines that guess again. The result is that Ostrokswki's method matches the efficiency of Halley's root finding method.
In version 1, the (error of the refined area/area of calculated result) is about 100! It seems that applying the refinement on two good estimates for the integral works well to enhance the final result only.
Namir
09-29-2016, 06:23 PM (This post was last modified: 09-29-2016 09:42 PM by Dieter.)
Post: #8
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Simpson revival
(09-29-2016 04:44 PM)Namir Wrote: So my version 1 of my Basic code is the way to go since it does not update the older area value in W.
IMHO the way to go is the approach in my HP41 program:
You calculate the Simpson approximation for n=2, 4, 8, ... intervals. After each step the improved value is calculated from the last two approximations. This value is returned to the user. If last two improved (!) approximations agree within the stated tolerance, the iteration exits.
In the given example two "improved" steps yield a similar accuracy as three "standard" Simpson steps, so the final result is obtained faster. If only one single iteration is saved, the total execution time is reduced by 50%.
(09-29-2016 04:44 PM)Namir Wrote: To be honest, I am a bit disappointed by your algorithm, since the general expectation for improving a calculated value in an iteration should be used in the next iteration and cause enhancing the result and/or reduce the number of iteration.
Just try it the way described above. ;-)
EDIT: Here is a short VBA program that implements the suggested method. | {
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"lm_q2_score": 0.8577681104440172,
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"openwebmath_score": 0.639910876750946,
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"url": "https://www.hpmuseum.org/forum/showthread.php?mode=linear&tid=6939&pid=61973"
} |
EDIT: Here is a short VBA program that implements the suggested method.
Code:
Function Simpson(a, b, Optional errlimit = 0.000001, Optional nmax = 2000) fab = f(a) + f(b) s2 = 0 s4 = f((a + b) / 2) simp_new = (fab + 4 * s4) * (b - a) / 6 improved_new = simp_new n = 4 Do h = (b - a) / n s2 = s2 + s4 s4 = 0 For i = 1 To n - 1 Step 2 s4 = s4 + f(a + i * h) Next simp_old = simp_new simp_new = (fab + 2 * s2 + 4 * s4) * h / 3 improved_old = improved_new improved_new = (16 * simp_new - simp_old) / 15 n = 2 * n Loop Until (Abs(improved_new - improved_old) <= errlimit) Or (n > nmax) Simpson = improved_new End Function Function f(x) f = 1 / x End Function
Dieter
09-29-2016, 10:27 PM (This post was last modified: 09-29-2016 10:30 PM by Namir.)
Post: #9
Namir Senior Member Posts: 813 Joined: Dec 2013
RE: Simpson revival
A modified version of the BASIC program works if we have two tests:
If Abs(Z - W) <= P Then
....
And
If Abs(Z - (16 * Z - W) / 15) <= P Then
...
I tested f(x)=1/X for integrals in [1,2], [1,10], and [1,100]. In each case the absolute value of the ratio of the error of the calculated integral divided by the error of the improved integral is in the order of 1000!
I suggest omitting the first test altogether!
Here is an updated Basic version that reflects Dieter's suggestion and my above comments: | {
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"lm_q1q2_score": 0.8489089497892652,
"lm_q2_score": 0.8577681104440172,
"openwebmath_perplexity": 529.6911219328429,
"openwebmath_score": 0.639910876750946,
"tags": null,
"url": "https://www.hpmuseum.org/forum/showthread.php?mode=linear&tid=6939&pid=61973"
} |
Here is an updated Basic version that reflects Dieter's suggestion and my above comments:
Code:
10 INPUT A 20 INPUT B 30 P=1E-5 'Expected tolerance 40 R=0: L=1: N=1: W=1E99 'Initializing 50 X=A: GOSUB 1000: T=Y 'Call Y=f(X) subroutine with X=A 60 X=B: GOSUB 1000: T=T+Y 'T=f(A)+f(B) now 70 N=N*2: H=(B-A)/N 'New step: Double the number of intervals 80 X=A+H: S=0 'Start with 1st new (odd) point 90 FOR I=1 TO L 'Calculate Sum of f(x) for new (odd) points 100 GOSUB 1000: S=S+Y 110 X=X+2*H 'Next new (odd) point 120 NEXT I 130 S=S*4: U=T+R/2 'Odd points have a coefficient of 4; Even points come from old odd points but with a coefficient of 2 instead of 4 140 Z=H*(S+U)/3 'New Integral value at this step 145 Y=(16*Z-W)/15 ' Calculate improved integral 150 IF ABS(Z-Y)<=P THEN PRINT "Integral value: ";Y: STOP 'OK with expected tolerance 160 W=Z: T=U: R=S: L=N 'Prepare new step 170 GOTO 70 'New step, please 1000 Y=f(X) 'Calculate f(X) subroutine 1010 RETURN
Namir
09-30-2016, 06:00 AM (This post was last modified: 09-30-2016 06:07 AM by Dieter.)
Post: #10
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Simpson revival
(09-29-2016 10:27 PM)Namir Wrote: A modified version of the BASIC program works if we have two tests:
If Abs(Z - W) <= P Then
This compares the old and new Simpson result. As you already said, this test is obsolete.
(09-29-2016 10:27 PM)Namir Wrote: If Abs(Z - (16 * Z - W) / 15) <= P Then
...
This compares the last Simpson result and the last improved approximation. Does this make sense?
(09-29-2016 10:27 PM)Namir Wrote: I suggest omitting the first test altogether!
The test should compare the last and previous improved approximations, cf. the posted VBA code. In your program that's the current and previous value of Y. | {
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"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9896718465668232,
"lm_q1q2_score": 0.8489089497892652,
"lm_q2_score": 0.8577681104440172,
"openwebmath_perplexity": 529.6911219328429,
"openwebmath_score": 0.639910876750946,
"tags": null,
"url": "https://www.hpmuseum.org/forum/showthread.php?mode=linear&tid=6939&pid=61973"
} |
Dieter
10-02-2016, 03:29 PM (This post was last modified: 10-02-2016 03:37 PM by Dieter.)
Post: #11
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Simpson revival
(09-30-2016 06:00 AM)I Wrote: The test should compare the last and previous improved approximations, cf. the posted VBA code. In your program that's the current and previous value of Y.
It looks like even this remaining error can be estimated and an even more accurate result can be calculated from the last two improved values. Here is a more structured experimental version.
Code:
Function Simpson(a, b, Optional errlimit = 0.000001, Optional nmax = 3000) fab = f(a) + f(b) s2 = 0 s4 = 0 simp_new = 0 improved_new = 0 n = 2 Do h = (b - a) / n s2 = s2 + s4 s4 = 0 For i = 1 To n - 1 Step 2 s4 = s4 + f(a + i * h) Next simp_old = simp_new improved_old = improved_new simp_new = (fab + 2 * s2 + 4 * s4) * h / 3 If n = 2 Then improved_new = simp_new converged = False Else improved_new = (16 * simp_new - simp_old) / 15 converged = Abs(improved_new - improved_old) < 63 * errlimit End If n = 2 * n Loop Until converged Or n > nmax Simpson = (64 * improved_new - improved_old) / 63 End Function Function f(x) f = 1 / x End Function
What do you think?
The following diagram may give an impression for the integral of f(x) = 1/x from a=1 to b=2. Here X is the number of iterations (i.e. n=2x intervals) and Y is the accuracy, i.e. the number of valid digits. The red line is the classic Simpson method, the blue one represents the "improved" method, and finally the green graph shows the further extrapolated result as shown in the last program. | {
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} |
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